गुणनखंड विधि से \(7x^2-9x+2=0\) के मूल क्या होंगे?
Using factorisation method, what will be the roots of \(7x^2-9x+2=0\)?
#quadratic
#factorisation
#fraction-roots
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A \(x=1,\frac{2}{7}\)
B \(x=-1,-\frac{2}{7}\)
C \(x=7,\frac{1}{2}\)
D \(x=2,\frac{1}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x=1,\frac{2}{7}\)
Step 1
Concept
(7x-2 -9x+2=(7x-2)(x-1)), so the roots are (1) and \(\frac{2}{7}\). In exams, solve each linear factor separately.
Step 2
Why this answer is correct
The correct answer is A. \(x=1,\frac{2}{7}\). (7x-2 -9x+2=(7x-2)(x-1)), so the roots are (1) and \(\frac{2}{7}\). In exams, solve each linear factor separately.
Step 3
Exam Tip
(7x-2 -9x+2=(7x-2)(x-1)), इसलिए मूल (1) और \(\frac{2}{7}\) हैं। परीक्षा में हर रैखिक गुणनखंड को अलग हल करें।
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\(10x^2-17x+3=0\) में मध्य पद का सही विभाजन कौनसा है?
What is the correct splitting of the middle term in \(10x^2-17x+3=0\)?
#quadratic
#middle-term-splitting
#ac-method
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A \(10x^2-15x-2x+3=0\)
B \(10x^2-12x-5x+3=0\)
C \(10x^2-10x-7x+3=0\)
D \(10x^2-20x+3x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(10x^2-15x-2x+3=0\)
Step 1
Concept
Here (ac=30) and (-15+(-2)=-17), so the middle term is (-15x-2x). In exams, check both sum and product.
Step 2
Why this answer is correct
The correct answer is A. \(10x^2-15x-2x+3=0\). Here (ac=30) and (-15+(-2)=-17), so the middle term is (-15x-2x). In exams, check both sum and product.
Step 3
Exam Tip
यहां (ac=30) और (-15+(-2)=-17), इसलिए मध्य पद (-15x-2x) होगा। परीक्षा में योग और गुणनफल दोनों जांचें।
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द्विघात सूत्र से \(x^2-10x+24=0\) के मूल क्या मिलेंगे?
Using the quadratic formula, what roots are obtained for \(x^2-10x+24=0\)?
#quadratic
#quadratic-formula
#roots
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A (x=4,6)
B (x=-4,-6)
C (x=2,12)
D (x=5,24)
Explanation opens after your attempt
Correct Answer
A. (x=4,6)
Step 1
Concept
Here (D=(-10)2 -4(1)(24)=4), so \(x=\frac{10\pm2}{2}\). In exams, keep the sign of (-b) correct.
Step 2
Why this answer is correct
The correct answer is A. (x=4,6). Here (D=(-10)2 -4(1)(24)=4), so \(x=\frac{10\pm2}{2}\). In exams, keep the sign of (-b) correct.
Step 3
Exam Tip
यहां (D=(-10)2 -4(1)(24)=4), इसलिए \(x=\frac{10\pm2}{2}\) मिलता है। परीक्षा में (-b) का चिन्ह सही रखें।
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\(x^2-10x+24=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-10x+24=0\) by completing the square?
#quadratic
#completing-square
#steps
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A ((x-5)2 =1)
B ((x+5)2 =1)
C ((x-10)2 =24)
D ((x-5)2 =24)
Explanation opens after your attempt
Correct Answer
A. ((x-5)2 =1)
Step 1
Concept
Adding (25) to \(x^2-10x=-24\) gives ((x-5)2 =1). In exams, add the same number to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x-5)2 =1). Adding (25) to \(x^2-10x=-24\) gives ((x-5)2 =1). In exams, add the same number to both sides.
Step 3
Exam Tip
\(x^2-10x=-24\) में (25) जोड़ने पर ((x-5)2 =1) मिलता है। परीक्षा में दोनों पक्षों में समान संख्या जोड़ें।
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\(12x^2+17x+5=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(12x^2+17x+5=0\) by factorisation?
#quadratic
#factorisation
#fraction-roots
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A \(x=-1,-\frac{5}{12}\)
B \(x=1,\frac{5}{12}\)
C \(x=-\frac{12}{5},-1\)
D \(x=-5,-\frac{1}{12}\)
Explanation opens after your attempt
Correct Answer
A. \(x=-1,-\frac{5}{12}\)
Step 1
Concept
(12x-2 +17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-1,-\frac{5}{12}\). (12x-2 +17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 3
Exam Tip
(12x-2 +17x+5=(12x+5)(x+1)), इसलिए मूल \(-\frac{5}{12}\) और (-1) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(4x^2-19x-5=0\) में मध्य पद का सही विभाजन कौनसा है?
What is the correct splitting of the middle term in \(4x^2-19x-5=0\)?
#quadratic
#middle-term-splitting
#signs
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A \(4x^2-20x+x-5=0\)
B \(4x^2-16x-3x-5=0\)
C \(4x^2-10x-9x-5=0\)
D \(4x^2-24x+5x-5=0\)
Explanation opens after your attempt
Correct Answer
A. \(4x^2-20x+x-5=0\)
Step 1
Concept
Here (ac=-20) and (-20+1=-19), so the middle term is (-20x+x). In exams, check the sign of (ac) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-20x+x-5=0\). Here (ac=-20) and (-20+1=-19), so the middle term is (-20x+x). In exams, check the sign of (ac) carefully.
Step 3
Exam Tip
यहां (ac=-20) और (-20+1=-19), इसलिए मध्य पद (-20x+x) होगा। परीक्षा में (ac) का संकेत ध्यान से देखें।
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\(4x^2-19x-5=0\) के मूल क्या हैं?
What are the roots of \(4x^2-19x-5=0\)?
#quadratic
#factorisation
#mixed-signs
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A \(x=5,-\frac{1}{4}\)
B \(x=-5,\frac{1}{4}\)
C (x=4,-5)
D \(x=\frac{5}{4},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=5,-\frac{1}{4}\)
Step 1
Concept
(4x-2 -19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=5,-\frac{1}{4}\). (4x-2 -19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 3
Exam Tip
(4x-2 -19x-5=(4x+1)(x-5)), इसलिए मूल (5) और \(-\frac{1}{4}\) हैं। परीक्षा में रैखिक गुणनखंडों के चिन्ह उलटकर मूल लिखें।
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यदि \(5x^2+9x=2\) है, तो मानक रूप में समीकरण क्या होगा?
If \(5x^2+9x=2\), what will be the equation in standard form?
#quadratic
#standard-form
#transposition
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A \(5x^2+9x-2=0\)
B \(5x^2+9x+2=0\)
C \(5x^2-9x-2=0\)
D \(9x^2+5x-2=0\)
Explanation opens after your attempt
Correct Answer
A. \(5x^2+9x-2=0\)
Step 1
Concept
Bringing (2) to the left gives \(5x^2+9x-2=0\). In exams, make standard form before solving.
Step 2
Why this answer is correct
The correct answer is A. \(5x^2+9x-2=0\). Bringing (2) to the left gives \(5x^2+9x-2=0\). In exams, make standard form before solving.
Step 3
Exam Tip
(2) को बाईं ओर लाने पर \(5x^2+9x-2=0\) मिलता है। परीक्षा में हल करने से पहले मानक रूप बनाएं।
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\(5x^2+9x-2=0\) के मूल क्या हैं?
What are the roots of \(5x^2+9x-2=0\)?
#quadratic
#standard-form
#factorisation
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A \(x=\frac{1}{5},-2\)
B \(x=-\frac{1}{5},2\)
C (x=5,-2)
D \(x=\frac{2}{5},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{5},-2\)
Step 1
Concept
(5x-2 +9x-2=(5x-1)(x+2)), so the roots are \(\frac{1}{5}\) and (-2). In exams, correct standard form is very important.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{5},-2\). (5x-2 +9x-2=(5x-1)(x+2)), so the roots are \(\frac{1}{5}\) and (-2). In exams, correct standard form is very important.
Step 3
Exam Tip
(5x-2 +9x-2=(5x-1)(x+2)), इसलिए मूल \(\frac{1}{5}\) और (-2) हैं। परीक्षा में सही मानक रूप बहुत जरूरी है।
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\(36x^2-60x+25=0\) किस पूर्ण वर्ग रूप में लिखा जाएगा?
In which perfect square form will \(36x^2-60x+25=0\) be written?
#quadratic
#perfect-square
#identity
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A ((6x-5)2 =0)
B ((6x+5)2 =0)
C ((36x-5)2 =0)
D ((x-5)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((6x-5)2 =0)
Step 1
Concept
(36x-2 -60x+25=(6x-5)2 ), so it is a perfect square. In exams, recognize the pattern ((a-b)2 ).
Step 2
Why this answer is correct
The correct answer is A. ((6x-5)2 =0). (36x-2 -60x+25=(6x-5)2 ), so it is a perfect square. In exams, recognize the pattern ((a-b)2 ).
Step 3
Exam Tip
(36x-2 -60x+25=(6x-5)2 ), इसलिए यह पूर्ण वर्ग है। परीक्षा में ((a-b)2 ) का पैटर्न पहचानें।
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\(36x^2-60x+25=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(36x^2-60x+25=0\)?
#quadratic
#repeated-root
#perfect-square
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A \(x=\frac{5}{6}\)
B \(x=-\frac{5}{6}\)
C \(x=\frac{6}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{6}\)
Step 1
Concept
((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{6}\). ((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((6x-5)2 =0), इसलिए (6x-5=0) और \(x=\frac{5}{6}\) है। परीक्षा में दोहराए हुए मूल को सही भिन्न में लिखें।
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\(8x^2-32x=0\) को हल करते समय कौनसी गलती नहीं करनी चाहिए?
Which mistake should be avoided while solving \(8x^2-32x=0\)?
#quadratic
#common-mistake
#zero-root
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A (x=0) को छोड़ना / Missing (x=0)
B (8x) सामान्य गुणनखंड लेना / Taking (8x) common
C (8x(x-4)=0) लिखना / Writing (8x(x-4)=0)
D (x=4) को मूल मानना / Taking (x=4) as a root
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ना / Missing (x=0)
Step 1
Concept
(8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ना / Missing (x=0). (8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 3
Exam Tip
(8x-2 -32x=8x(x-4)), इसलिए (x=0) और (x=4) दोनों मूल हैं। परीक्षा में चर से भाग देने पर (x=0) छूट सकता है।
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द्विघात सूत्र से \(5x^2-10x-3=0\) के लिए (D) का मान क्या है?
Using the quadratic formula setup, what is the value of (D) for \(5x^2-10x-3=0\)?
#quadratic
#discriminant
#calculation
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A (160)
B (100)
C (60)
D (40)
Explanation opens after your attempt
Step 1
Concept
Here (D=(-10)2 -4(5)(-3)=160). In exams, a negative (c) makes the second term add.
Step 2
Why this answer is correct
The correct answer is A. (160). Here (D=(-10)2 -4(5)(-3)=160). In exams, a negative (c) makes the second term add.
Step 3
Exam Tip
यहां (D=(-10)2 -4(5)(-3)=160) है। परीक्षा में ऋणात्मक (c) के कारण दूसरा पद जुड़ता है।
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\(5x^2-10x-3=0\) के मूलों का सही रूप क्या है?
What is the correct form of the roots of \(5x^2-10x-3=0\)?
#quadratic
#quadratic-formula
#irrational-roots
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A \(x=1\pm\frac{2\sqrt{10}}{5}\)
B \(x=1\pm2\sqrt{10}\)
C \(x=\frac{1\pm2\sqrt{10}}{5}\)
D \(x=5\pm2\sqrt{10}\)
Explanation opens after your attempt
Correct Answer
A. \(x=1\pm\frac{2\sqrt{10}}{5}\)
Step 1
Concept
The formula gives \(x=\frac{10\pm\sqrt{160}}{10}=1\pm\frac{2\sqrt{10}}{5}\). In exams, simplify \(\sqrt{160}=4\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=1\pm\frac{2\sqrt{10}}{5}\). The formula gives \(x=\frac{10\pm\sqrt{160}}{10}=1\pm\frac{2\sqrt{10}}{5}\). In exams, simplify \(\sqrt{160}=4\sqrt{10}\).
Step 3
Exam Tip
सूत्र से \(x=\frac{10\pm\sqrt{160}}{10}=1\pm\frac{2\sqrt{10}}{5}\) मिलता है। परीक्षा में \(\sqrt{160}=4\sqrt{10}\) सरल करें।
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यदि \(x^2-16x+k=0\) के मूल समान हों, तो (k) का मान क्या होगा?
If \(x^2-16x+k=0\) has equal roots, what is the value of (k)?
#quadratic
#discriminant
#parameter
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A (64)
B (16)
C (32)
D (256)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0), so (256-4k=0) and (k=64). In exams, equal roots indicate (D=0).
Step 2
Why this answer is correct
The correct answer is A. (64). For equal roots, (D=0), so (256-4k=0) and (k=64). In exams, equal roots indicate (D=0).
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए (256-4k=0) और (k=64) है। परीक्षा में समान मूल का संकेत (D=0) है।
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यदि \(x^2+kx+36=0\) के समान मूल हैं और (k>0), तो (k) क्या होगा?
If \(x^2+kx+36=0\) has equal roots and (k>0), what is (k)?
#quadratic
#discriminant
#parameter
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A (12)
B (-12)
C (6)
D (36)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(k^2-144=0\), so \(k=\pm12\), and (k>0) gives (k=12). In exams, apply the given condition.
Step 2
Why this answer is correct
The correct answer is A. (12). For equal roots, \(k^2-144=0\), so \(k=\pm12\), and (k>0) gives (k=12). In exams, apply the given condition.
Step 3
Exam Tip
समान मूलों के लिए \(k^2-144=0\), इसलिए \(k=\pm12\) और (k>0) से (k=12) है। परीक्षा में दी गई शर्त जरूर लगाएं।
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\(x^2-11x+30=0\) और \(x^2-13x+42=0\) में कौनसा मूल समान है?
Which root is common to \(x^2-11x+30=0\) and \(x^2-13x+42=0\)?
#quadratic
#common-root
#factorisation
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A (x=6)
B (x=5)
C (x=7)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (5,6) और दूसरे के मूल (6,7) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
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\(49x^2-42x+9=0\) को किस पूर्ण वर्ग रूप में लिखा जाएगा?
In which perfect square form will \(49x^2-42x+9=0\) be written?
#quadratic
#perfect-square
#identity
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A ((7x-3)2 =0)
B ((7x+3)2 =0)
C ((49x-3)2 =0)
D ((x-3)2 =0)
Explanation opens after your attempt
Correct Answer
A. ((7x-3)2 =0)
Step 1
Concept
(49x-2 -42x+9=(7x-3)2 ), so it is a perfect square. In exams, check \(2\cdot7x\cdot3=42x\).
Step 2
Why this answer is correct
The correct answer is A. ((7x-3)2 =0). (49x-2 -42x+9=(7x-3)2 ), so it is a perfect square. In exams, check \(2\cdot7x\cdot3=42x\).
Step 3
Exam Tip
(49x-2 -42x+9=(7x-3)2 ), इसलिए यह पूर्ण वर्ग है। परीक्षा में \(2\cdot7x\cdot3=42x\) जांचें।
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\(49x^2-42x+9=0\) का मूल क्या है?
What is the root of \(49x^2-42x+9=0\)?
#quadratic
#repeated-root
#perfect-square
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A \(x=\frac{3}{7}\)
B \(x=-\frac{3}{7}\)
C \(x=\frac{7}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{7}\)
Step 1
Concept
((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{7}\). ((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((7x-3)2 =0), इसलिए (7x-3=0) और \(x=\frac{3}{7}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(3x^2+2=11x\) को मानक रूप में लिखने पर क्या मिलेगा?
What is obtained when \(3x^2+2=11x\) is written in standard form?
#quadratic
#standard-form
#transposition
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A \(3x^2-11x+2=0\)
B \(3x^2+11x+2=0\)
C \(3x^2-11x-2=0\)
D \(3x^2+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-11x+2=0\)
Step 1
Concept
Bringing (11x) to the left gives \(3x^2-11x+2=0\). In exams, bring all terms to one side.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-11x+2=0\). Bringing (11x) to the left gives \(3x^2-11x+2=0\). In exams, bring all terms to one side.
Step 3
Exam Tip
(11x) को बाईं ओर लाने पर \(3x^2-11x+2=0\) बनता है। परीक्षा में सभी पदों को एक तरफ लाएं।
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\(3x^2-11x+2=0\) का विविक्तकर (D) क्या है?
What is the discriminant (D) of \(3x^2-11x+2=0\)?
#quadratic
#discriminant
#calculation
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A (97)
B (121)
C (73)
D (105)
Explanation opens after your attempt
Step 1
Concept
Here (D=(-11)2 -4(3)(2)=97). In exams, identify (a,b,c) correctly while finding (D).
Step 2
Why this answer is correct
The correct answer is A. (97). Here (D=(-11)2 -4(3)(2)=97). In exams, identify (a,b,c) correctly while finding (D).
Step 3
Exam Tip
यहां (D=(-11)2 -4(3)(2)=97) है। परीक्षा में (a,b,c) सही पहचानकर (D) निकालें।
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\(x^2-18x+45=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-18x+45=0\) by completing square?
#quadratic
#completing-square
#steps
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A ((x-9)2 =36)
B ((x+9)2 =36)
C ((x-18)2 =45)
D ((x-9)2 =45)
Explanation opens after your attempt
Correct Answer
A. ((x-9)2 =36)
Step 1
Concept
Adding (81) to \(x^2-18x=-45\) gives ((x-9)2 =36). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x-9)2 =36). Adding (81) to \(x^2-18x=-45\) gives ((x-9)2 =36). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2-18x=-45\) में (81) जोड़ने पर ((x-9)2 =36) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(x^2-18x+45=0\) के मूल क्या हैं?
What are the roots of \(x^2-18x+45=0\)?
#quadratic
#completing-square
#roots
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A (x=3,15)
B (x=-3,-15)
C (x=6,12)
D (x=9,36)
Explanation opens after your attempt
Correct Answer
A. (x=3,15)
Step 1
Concept
Since ((x-9)2 =36), \(x-9=\pm6\), so (x=3,15). In exams, use \(\pm\) to find both roots.
Step 2
Why this answer is correct
The correct answer is A. (x=3,15). Since ((x-9)2 =36), \(x-9=\pm6\), so (x=3,15). In exams, use \(\pm\) to find both roots.
Step 3
Exam Tip
((x-9)2 =36), इसलिए \(x-9=\pm6\) और (x=3,15) हैं। परीक्षा में \(\pm\) से दोनों मूल निकालें।
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यदि \(x^2+px+42=0\) के मूल (-6) और (-7) हैं, तो (p) क्या है?
If the roots of \(x^2+px+42=0\) are (-6) and (-7), what is (p)?
#quadratic
#roots-to-equation
#parameter
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A (13)
B (-13)
C (42)
D (1)
Explanation opens after your attempt
Step 1
Concept
((x+6)(x+7)=x-2 +13x+42), so (p=13). In exams, form factors from given roots.
Step 2
Why this answer is correct
The correct answer is A. (13). ((x+6)(x+7)=x-2 +13x+42), so (p=13). In exams, form factors from given roots.
Step 3
Exam Tip
((x+6)(x+7)=x-2 +13x+42), इसलिए (p=13) है। परीक्षा में दिए मूलों से गुणनखंड बनाएं।
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यदि (4) और (9) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (4) and (9) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
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A \(x^2-13x+36=0\)
B \(x^2+13x+36=0\)
C \(x^2-36x+13=0\)
D \(x^2+36x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+36=0\)
Step 1
Concept
If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+36=0\). If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (4) और (9) हों तो ((x-4)(x-9)=0), यानी \(x^2-13x+36=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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\(5x^2+16x+3=0\) को हल करने पर मूल क्या होंगे?
What will be the roots after solving \(5x^2+16x+3=0\)?
#quadratic
#factorisation
#fraction-roots
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A \(x=-3,-\frac{1}{5}\)
B \(x=3,\frac{1}{5}\)
C \(x=-5,-\frac{3}{1}\)
D \(x=-1,-\frac{3}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x=-3,-\frac{1}{5}\)
Step 1
Concept
(5x-2 +16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-3,-\frac{1}{5}\). (5x-2 +16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 3
Exam Tip
(5x-2 +16x+3=(5x+1)(x+3)), इसलिए मूल \(-\frac{1}{5}\) और (-3) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-4x-32=0\) के लिए सही कथन कौनसा है?
Which statement is correct for \(x^2-4x-32=0\)?
#quadratic
#roots
#mixed-signs
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A मूल (8) और (-4) हैं / The roots are (8) and (-4)
B मूल (-8) और (4) हैं / The roots are (-8) and (4)
C मूल (4) और (32) हैं / The roots are (4) and (32)
D मूल (-4) और (-32) हैं / The roots are (-4) and (-32)
Explanation opens after your attempt
Correct Answer
A. मूल (8) और (-4) हैं / The roots are (8) and (-4)
Step 1
Concept
(x-2 -4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.
Step 2
Why this answer is correct
The correct answer is A. मूल (8) और (-4) हैं / The roots are (8) and (-4). (x-2 -4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.
Step 3
Exam Tip
(x-2 -4x-32=(x-8)(x+4)), इसलिए मूल (8) और (-4) हैं। परीक्षा में बड़ा मान मध्य पद का चिन्ह तय करता है।
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द्विघात सूत्र में (a=1,b=-14,c=45) रखने पर मूल क्या होंगे?
What roots are obtained by putting (a=1,b=-14,c=45) in the quadratic formula?
#quadratic
#quadratic-formula
#substitution
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A (x=5,9)
B (x=-5,-9)
C (x=3,15)
D (x=7,8)
Explanation opens after your attempt
Correct Answer
A. (x=5,9)
Step 1
Concept
(D=(-14)2 -4(1)(45)=16), so \(x=\frac{14\pm4}{2}\) gives (5) and (9). In exams, keep the sign of (b) correct in the formula.
Step 2
Why this answer is correct
The correct answer is A. (x=5,9). (D=(-14)2 -4(1)(45)=16), so \(x=\frac{14\pm4}{2}\) gives (5) and (9). In exams, keep the sign of (b) correct in the formula.
Step 3
Exam Tip
(D=(-14)2 -4(1)(45)=16), इसलिए \(x=\frac{14\pm4}{2}\) से (5) और (9) मिलते हैं। परीक्षा में सूत्र में (b) का चिन्ह सही रखें।
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\(x^2-20x+96=0\) में कौनसी संख्या जोड़ी मध्य पद बनाने में मदद करेगी?
Which number pair helps in making the middle term in \(x^2-20x+96=0\)?
#quadratic
#middle-term-splitting
#signs
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A (-8) और (-12) / (-8) and (-12)
B (8) और (12) / (8) and (12)
C (-6) और (-16) / (-6) and (-16)
D (6) और (16) / (6) and (16)
Explanation opens after your attempt
Correct Answer
A. (-8) और (-12) / (-8) and (-12)
Step 1
Concept
(-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.
Step 2
Why this answer is correct
The correct answer is A. (-8) और (-12) / (-8) and (-12). (-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.
Step 3
Exam Tip
(-8+(-12)=-20) और ((-8)(-12)=96), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक होती हैं।
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\(x^2-20x+96=0\) के मूल क्या हैं?
What are the roots of \(x^2-20x+96=0\)?
#quadratic
#roots
#factorisation
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A (x=8,12)
B (x=-8,-12)
C (x=6,16)
D (x=-6,-16)
Explanation opens after your attempt
Correct Answer
A. (x=8,12)
Step 1
Concept
(x-2 -20x+96=(x-8)(x-12)), so the roots are (8) and (12). In exams, change signs while writing roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=8,12). (x-2 -20x+96=(x-8)(x-12)), so the roots are (8) and (12). In exams, change signs while writing roots from factors.
Step 3
Exam Tip
(x-2 -20x+96=(x-8)(x-12)), इसलिए मूल (8) और (12) हैं। परीक्षा में गुणनखंड से मूल लिखते समय चिन्ह बदलें।
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\(11x^2=77x\) को हल करने का सही तरीका क्या है?
What is the correct way to solve \(11x^2=77x\)?
#quadratic
#common-factor
#zero-product
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A (11x(x-7)=0) लिखना / Write (11x(x-7)=0)
B (x=7) ही लिखना / Write only (x=7)
C (11x=77) लिखना / Write (11x=77)
D \(x^2=7x\) के बाद (x=7) ही लेना / After \(x^2=7x\), take only (x=7)
Explanation opens after your attempt
Correct Answer
A. (11x(x-7)=0) लिखना / Write (11x(x-7)=0)
Step 1
Concept
From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (11x(x-7)=0) लिखना / Write (11x(x-7)=0). From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(11x^2-77x=0\) से (11x(x-7)=0), इसलिए (x=0) और (x=7) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
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\(x^2+8x-33=0\) को पूर्ण वर्ग विधि से हल करने में सही चरण कौनसा है?
Which step is correct in solving \(x^2+8x-33=0\) by completing square?
#quadratic
#completing-square
#steps
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A ((x+4)2 =49)
B ((x-4)2 =49)
C ((x+8)2 =33)
D ((x+4)2 =33)
Explanation opens after your attempt
Correct Answer
A. ((x+4)2 =49)
Step 1
Concept
Adding (16) to \(x^2+8x=33\) gives ((x+4)2 =49). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x+4)2 =49). Adding (16) to \(x^2+8x=33\) gives ((x+4)2 =49). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2+8x=33\) में (16) जोड़ने पर ((x+4)2 =49) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(x^2+8x-33=0\) के मूल क्या हैं?
What are the roots of \(x^2+8x-33=0\)?
#quadratic
#completing-square
#roots
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A (x=3,-11)
B (x=-3,11)
C (x=7,-7)
D (x=8,-33)
Explanation opens after your attempt
Correct Answer
A. (x=3,-11)
Step 1
Concept
Since ((x+4)2 =49), \(x+4=\pm7\), so (x=3,-11). In exams, use \(\pm\) to get both answers.
Step 2
Why this answer is correct
The correct answer is A. (x=3,-11). Since ((x+4)2 =49), \(x+4=\pm7\), so (x=3,-11). In exams, use \(\pm\) to get both answers.
Step 3
Exam Tip
((x+4)2 =49), इसलिए \(x+4=\pm7\) और (x=3,-11) हैं। परीक्षा में \(\pm\) से दोनों उत्तर निकालें।
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\(8x^2-2x-3=0\) में मध्य पद का सही विभाजन कौनसा है?
What is the correct splitting of the middle term in \(8x^2-2x-3=0\)?
#quadratic
#middle-term-splitting
#ac-method
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A \(8x^2+4x-6x-3=0\)
B \(8x^2+2x-4x-3=0\)
C \(8x^2+6x-8x-3=0\)
D \(8x^2-4x+2x-3=0\)
Explanation opens after your attempt
Correct Answer
A. \(8x^2+4x-6x-3=0\)
Step 1
Concept
(ac=-24) and (4+(-6)=-2), so (-2x) is split as (4x-6x). In exams, check the sign of (ac) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(8x^2+4x-6x-3=0\). (ac=-24) and (4+(-6)=-2), so (-2x) is split as (4x-6x). In exams, check the sign of (ac) carefully.
Step 3
Exam Tip
(ac=-24) और (4+(-6)=-2), इसलिए (-2x) को (4x-6x) में तोड़ते हैं। परीक्षा में (ac) का संकेत ध्यान से देखें।
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\(8x^2-2x-3=0\) के मूल क्या हैं?
What are the roots of \(8x^2-2x-3=0\)?
#quadratic
#factorisation
#roots
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A \(x=\frac{3}{4},-\frac{1}{2}\)
B \(x=-\frac{3}{4},\frac{1}{2}\)
C \(x=2,-\frac{3}{8}\)
D \(x=\frac{1}{4},-3\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{4},-\frac{1}{2}\)
Step 1
Concept
(8x-2 -2x-3=(4x-3)(2x+1)), so the roots are \(\frac{3}{4}\) and \(-\frac{1}{2}\). In exams, solve both linear factors carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{4},-\frac{1}{2}\). (8x-2 -2x-3=(4x-3)(2x+1)), so the roots are \(\frac{3}{4}\) and \(-\frac{1}{2}\). In exams, solve both linear factors carefully.
Step 3
Exam Tip
(8x-2 -2x-3=(4x-3)(2x+1)), इसलिए मूल \(\frac{3}{4}\) और \(-\frac{1}{2}\) हैं। परीक्षा में दोनों रैखिक गुणनखंड सावधानी से हल करें।
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यदि (D=0) और (a=4,b=-16), तो समान मूल क्या होगा?
If (D=0) and (a=4,b=-16), what will be the equal root?
#quadratic
#equal-roots
#formula
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A (x=2)
B (x=-2)
C (x=4)
D (x=-4)
Explanation opens after your attempt
Step 1
Concept
The equal root is \(x=\frac{-b}{2a}\), so \(x=\frac{16}{8}=2\). In exams, this short formula is useful when (D=0).
Step 2
Why this answer is correct
The correct answer is A. (x=2). The equal root is \(x=\frac{-b}{2a}\), so \(x=\frac{16}{8}=2\). In exams, this short formula is useful when (D=0).
Step 3
Exam Tip
समान मूल \(x=\frac{-b}{2a}\) होता है, इसलिए \(x=\frac{16}{8}=2\) है। परीक्षा में (D=0) पर यह छोटा सूत्र उपयोगी है।
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\(x^2-6x+18=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about real roots of \(x^2-6x+18=0\)?
#quadratic
#discriminant
#no-real-roots
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A कोई वास्तविक मूल नहीं / No real roots
B दो समान वास्तविक मूल / Two equal real roots
C दो अलग वास्तविक मूल / Two distinct real roots
D एक मूल (0) / One root (0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
Here (D=(-6)2 -4(1)(18)=-36<0), so there are no real roots. In exams, (D<0) means no real solution.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. Here (D=(-6)2 -4(1)(18)=-36<0), so there are no real roots. In exams, (D<0) means no real solution.
Step 3
Exam Tip
यहां (D=(-6)2 -4(1)(18)=-36<0), इसलिए वास्तविक मूल नहीं हैं। परीक्षा में (D<0) होने पर वास्तविक हल नहीं मिलता।
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\(x^2-6x+18=0\) में पूर्ण वर्ग बनाने पर कौनसा रूप बनेगा?
What form is obtained by completing the square in \(x^2-6x+18=0\)?
#quadratic
#completing-square
#no-real-roots
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A ((x-3)2 +9=0)
B ((x+3)2 +9=0)
C ((x-3)2 -9=0)
D ((x+6)2 +3=0)
Explanation opens after your attempt
Correct Answer
A. ((x-3)2 +9=0)
Step 1
Concept
(x-2 -6x+18=(x-3)2 +9), so no real roots are obtained. In exams, use completed square form to understand the nature of roots.
Step 2
Why this answer is correct
The correct answer is A. ((x-3)2 +9=0). (x-2 -6x+18=(x-3)2 +9), so no real roots are obtained. In exams, use completed square form to understand the nature of roots.
Step 3
Exam Tip
(x-2 -6x+18=(x-3)2 +9), इसलिए वास्तविक मूल नहीं मिलते। परीक्षा में पूर्ण वर्ग रूप से भी मूलों की प्रकृति समझें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
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A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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((x-7)2 =11) को हल करने पर (x) का मान क्या होगा?
Solving ((x-7)2 =11), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
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A \(x=7\pm\sqrt{11}\)
B \(x=-7\pm\sqrt{11}\)
C \(x=7\pm11\)
D \(x=\sqrt{7}\pm11\)
Explanation opens after your attempt
Correct Answer
A. \(x=7\pm\sqrt{11}\)
Step 1
Concept
\(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=7\pm\sqrt{11}\). \(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x-7=\pm\sqrt{11}\), इसलिए \(x=7\pm\sqrt{11}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(x^2+14x+48=0\) के लिए सही हल कौनसा है?
Which is the correct solution for \(x^2+14x+48=0\)?
#quadratic
#factorisation
#sign-concept
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A (x=-6,-8)
B (x=6,8)
C (x=-14,-48)
D (x=14,48)
Explanation opens after your attempt
Correct Answer
A. (x=-6,-8)
Step 1
Concept
(x-2 +14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-6,-8). (x-2 +14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 3
Exam Tip
(x-2 +14x+48=(x+6)(x+8)), इसलिए (x=-6,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं।
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\(11x^2+12x+1=0\) में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for \(11x^2+12x+1=0\)?
#quadratic
#factorisation
#verification
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A ((11x+1)(x+1)=0)
B ((11x-1)(x-1)=0)
C ((x+11)(x+1)=0)
D ((11x+12)(x+1)=0)
Explanation opens after your attempt
Correct Answer
A. ((11x+1)(x+1)=0)
Step 1
Concept
((11x+1)(x+1)=11x-2 +12x+1), so it is correct. In exams, verify the factors by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((11x+1)(x+1)=0). ((11x+1)(x+1)=11x-2 +12x+1), so it is correct. In exams, verify the factors by expanding.
Step 3
Exam Tip
((11x+1)(x+1)=11x-2 +12x+1), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(11x^2+12x+1=0\) के मूल क्या होंगे?
What will be the roots of \(11x^2+12x+1=0\)?
#quadratic
#fraction-roots
#zero-product
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A \(x=-\frac{1}{11},-1\)
B \(x=\frac{1}{11},1\)
C (x=-11,-1)
D (x=11,1)
Explanation opens after your attempt
Correct Answer
A. \(x=-\frac{1}{11},-1\)
Step 1
Concept
((11x+1)(x+1)=0), so \(x=-\frac{1}{11}\) and (-1). In exams, ((11x+1)=0) gives \(-\frac{1}{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{1}{11},-1\). ((11x+1)(x+1)=0), so \(x=-\frac{1}{11}\) and (-1). In exams, ((11x+1)=0) gives \(-\frac{1}{11}\).
Step 3
Exam Tip
((11x+1)(x+1)=0), इसलिए \(x=-\frac{1}{11}\) और (-1) हैं। परीक्षा में ((11x+1)=0) से \(-\frac{1}{11}\) मिलता है।
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\(x^2+3x-3=0\) के मूल द्विघात सूत्र से क्या हैं?
What are the roots of \(x^2+3x-3=0\) by the quadratic formula?
#quadratic
#quadratic-formula
#irrational-roots
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A \(x=\frac{-3\pm\sqrt{21}}{2}\)
B \(x=\frac{3\pm\sqrt{21}}{2}\)
C \(x=-3\pm\sqrt{21}\)
D \(x=\frac{-3\pm\sqrt{9}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{-3\pm\sqrt{21}}{2}\)
Step 1
Concept
Here (D=32 -4(1)(-3)=21), so \(x=\frac{-3\pm\sqrt{21}}{2}\). In exams, keep the sign of (c=-3) correct.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{-3\pm\sqrt{21}}{2}\). Here (D=32 -4(1)(-3)=21), so \(x=\frac{-3\pm\sqrt{21}}{2}\). In exams, keep the sign of (c=-3) correct.
Step 3
Exam Tip
यहां (D=32 -4(1)(-3)=21), इसलिए \(x=\frac{-3\pm\sqrt{21}}{2}\) है। परीक्षा में (c=-3) का संकेत सही रखें।
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\(4x^2-12x-7=0\) में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for \(4x^2-12x-7=0\)?
#quadratic
#factorisation
#verification
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A ((2x+1)(2x-7)=0)
B ((2x-1)(2x+7)=0)
C ((4x-7)(x+1)=0)
D ((x+7)(4x-1)=0)
Explanation opens after your attempt
Correct Answer
A. ((2x+1)(2x-7)=0)
Step 1
Concept
((2x+1)(2x-7)=4x-2 -12x-7), so this is the correct factorised form. In exams, check the answer by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((2x+1)(2x-7)=0). ((2x+1)(2x-7)=4x-2 -12x-7), so this is the correct factorised form. In exams, check the answer by expanding.
Step 3
Exam Tip
((2x+1)(2x-7)=4x-2 -12x-7), इसलिए यह सही गुणनखंड रूप है। परीक्षा में विस्तार करके उत्तर जांचें।
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\(4x^2-12x-7=0\) के मूल क्या हैं?
What are the roots of \(4x^2-12x-7=0\)?
#quadratic
#roots
#factorisation
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A \(x=\frac{7}{2},-\frac{1}{2}\)
B \(x=-\frac{7}{2},\frac{1}{2}\)
C (x=2,-7)
D \(x=\frac{7}{4},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{7}{2},-\frac{1}{2}\)
Step 1
Concept
((2x+1)(2x-7)=0), so \(x=-\frac{1}{2}\) and \(\frac{7}{2}\). In exams, change signs while writing roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{7}{2},-\frac{1}{2}\). ((2x+1)(2x-7)=0), so \(x=-\frac{1}{2}\) and \(\frac{7}{2}\). In exams, change signs while writing roots.
Step 3
Exam Tip
((2x+1)(2x-7)=0), इसलिए \(x=-\frac{1}{2}\) और \(\frac{7}{2}\) हैं। परीक्षा में संकेत बदलकर मूल लिखें।
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यदि किसी छात्र ने \(x^2=81\) से केवल (x=9) लिखा, तो सही सुधार क्या है?
If a student wrote only (x=9) from \(x^2=81\), what is the correct correction?
#quadratic
#square-root-method
#common-mistake
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A \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\)
B (x=81) लिखना चाहिए / One should write (x=81)
C (x=-81) लिखना चाहिए / One should write (x=-81)
D \(x=\pm81\) लिखना चाहिए / One should write \(x=\pm81\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\)
Step 1
Concept
From \(x^2=81\), \(x=\pm\sqrt{81}=\pm9\). In exams, both signs are necessary in the square root method.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\). From \(x^2=81\), \(x=\pm\sqrt{81}=\pm9\). In exams, both signs are necessary in the square root method.
Step 3
Exam Tip
\(x^2=81\) से \(x=\pm\sqrt{81}=\pm9\) मिलता है। परीक्षा में वर्गमूल विधि में दोनों चिन्ह अनिवार्य हैं।
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यदि \(x^2-24x+108=0\) को पूर्ण वर्ग विधि से हल किया जाए, तो सही मध्य चरण कौनसा है?
If \(x^2-24x+108=0\) is solved by completing the square, which middle step is correct?
#quadratic
#completing-square
#steps
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A ((x-12)2 =36)
B ((x+12)2 =36)
C ((x-24)2 =108)
D ((x-12)2 =108)
Explanation opens after your attempt
Correct Answer
A. ((x-12)2 =36)
Step 1
Concept
Adding (144) to \(x^2-24x=-108\) gives ((x-12)2 =36). In exams, add the square of half the coefficient to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x-12)2 =36). Adding (144) to \(x^2-24x=-108\) gives ((x-12)2 =36). In exams, add the square of half the coefficient to both sides.
Step 3
Exam Tip
\(x^2-24x=-108\) में (144) जोड़ने पर ((x-12)2 =36) मिलता है। परीक्षा में आधे गुणांक का वर्ग दोनों पक्षों में जोड़ें।
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\(x^2-24x+108=0\) के मूल पूर्ण वर्ग विधि से क्या होंगे?
What roots are obtained for \(x^2-24x+108=0\) by completing the square method?
#quadratic
#completing-square
#roots
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A (x=6,18)
B (x=-6,-18)
C (x=12,6)
D (x=24,108)
Explanation opens after your attempt
Correct Answer
A. (x=6,18)
Step 1
Concept
Since ((x-12)2 =36), \(x-12=\pm6\), so (x=6,18). In exams, use \(\pm\) to find both roots.
Step 2
Why this answer is correct
The correct answer is A. (x=6,18). Since ((x-12)2 =36), \(x-12=\pm6\), so (x=6,18). In exams, use \(\pm\) to find both roots.
Step 3
Exam Tip
((x-12)2 =36), इसलिए \(x-12=\pm6\) और (x=6,18) हैं। परीक्षा में \(\pm\) से दोनों मूल निकालें।
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यदि \(x^2-2kx+25=0\) के समान मूल हों और (k>0), तो (k) का मान क्या होगा?
If \(x^2-2kx+25=0\) has equal roots and (k>0), what is the value of (k)?
#quadratic
#discriminant
#parameter
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A (5)
B (-5)
C (10)
D (25)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0), so ((-2k)2 -4(1)(25)=0) gives (k=5). In exams, apply the condition (k>0).
Step 2
Why this answer is correct
The correct answer is A. (5). For equal roots, (D=0), so ((-2k)2 -4(1)(25)=0) gives (k=5). In exams, apply the condition (k>0).
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-2k)2 -4(1)(25)=0) से (k=5) मिलता है। परीक्षा में (k>0) वाली शर्त जरूर लगाएं।
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