A. (\left\(3,0\right\)) और (\left\(0,-8\right\))/(\left\(3,0\right\)) and (\left\(0,-8\right\))
Step 1
Concept
At (y=0), (x=3), and at (x=0), (y=-8). Plot the negative intercept in the correct direction on the graph.
Step 2
Why this answer is correct
The correct answer is A. (\left\(3,0\right\)) और (\left\(0,-8\right\)) / (\left\(3,0\right\)) and (\left\(0,-8\right\)). At (y=0), (x=3), and at (x=0), (y=-8). Plot the negative intercept in the correct direction on the graph.
Step 3
Exam Tip
(y=0) पर (x=3) और (x=0) पर (y=-8)। ऋण अवरोध को ग्राफ में सही दिशा में अंकित करें।
A. (\left\(0,-4\right\)) और (\left\(16,0\right\))/(\left\(0,-4\right\)) and (\left\(16,0\right\))
Step 1
Concept
At (x=0), (y=-4), and at (y=0), (x=16). While finding intercepts, note which variable is kept zero.
Step 2
Why this answer is correct
The correct answer is A. (\left\(0,-4\right\)) और (\left\(16,0\right\)) / (\left\(0,-4\right\)) and (\left\(16,0\right\)). At (x=0), (y=-4), and at (y=0), (x=16). While finding intercepts, note which variable is kept zero.
Step 3
Exam Tip
(x=0) पर (y=-4) और (y=0) पर (x=16)। अवरोध निकालते समय कौन-सा चर शून्य रखा है, यह ध्यान रखें।
A. (\left\(5,0\right\)) और (\left\(0,-9\right\))/(\left\(5,0\right\)) and (\left\(0,-9\right\))
Step 1
Concept
At (y=0), (x=5), and at (x=0), (y=-9). Plot the negative intercept in the correct direction.
Step 2
Why this answer is correct
The correct answer is A. (\left\(5,0\right\)) और (\left\(0,-9\right\)) / (\left\(5,0\right\)) and (\left\(0,-9\right\)). At (y=0), (x=5), and at (x=0), (y=-9). Plot the negative intercept in the correct direction.
Step 3
Exam Tip
(y=0) पर (x=5) और (x=0) पर (y=-9)। ऋण अवरोध को सही दिशा में अंकित करें।
A. (\left\(0,-4\right\)) और (\left\(12,0\right\))/(\left\(0,-4\right\)) and (\left\(12,0\right\))
Step 1
Concept
At (x=0), (y=-4), and at (y=0), (x=12). While finding intercepts, note which variable is kept zero.
Step 2
Why this answer is correct
The correct answer is A. (\left\(0,-4\right\)) और (\left\(12,0\right\)) / (\left\(0,-4\right\)) and (\left\(12,0\right\)). At (x=0), (y=-4), and at (y=0), (x=12). While finding intercepts, note which variable is kept zero.
Step 3
Exam Tip
(x=0) पर (y=-4) और (y=0) पर (x=12)। अवरोध निकालते समय कौन-सा चर शून्य रखा है, यह ध्यान रखें।
A. (\left\(4,0\right\)) और (\left\(0,-7\right\))/(\left\(4,0\right\)) and (\left\(0,-7\right\))
Step 1
Concept
At (y=0), (x=4), and at (x=0), (y=-7). Plot the negative intercept in the correct direction.
Step 2
Why this answer is correct
The correct answer is A. (\left\(4,0\right\)) और (\left\(0,-7\right\)) / (\left\(4,0\right\)) and (\left\(0,-7\right\)). At (y=0), (x=4), and at (x=0), (y=-7). Plot the negative intercept in the correct direction.
Step 3
Exam Tip
(y=0) पर (x=4) और (x=0) पर (y=-7)। ऋण अवरोध को सही दिशा में अंकित करें।
A. (\left\(10,0\right\)) और (\left\(0,6\right\))/(\left\(10,0\right\)) and (\left\(0,6\right\))
Step 1
Concept
Putting (y=0) gives (x=10), and putting (x=0) gives (y=6). Intercepts make line drawing easier.
Step 2
Why this answer is correct
The correct answer is A. (\left\(10,0\right\)) और (\left\(0,6\right\)) / (\left\(10,0\right\)) and (\left\(0,6\right\)). Putting (y=0) gives (x=10), and putting (x=0) gives (y=6). Intercepts make line drawing easier.
Step 3
Exam Tip
(y=0) रखने पर (x=10) और (x=0) रखने पर (y=6)। अवरोधों से रेखा खींचना आसान होता है।
A. ( (10,0) ) और ( (0,4) )/( (10,0) ) and ( (0,4) )
Step 1
Concept
Putting (y=0) gives (x=10), and putting (x=0) gives (y=4). Intercepts make the graph quick and clear.
Step 2
Why this answer is correct
The correct answer is A. ( (10,0) ) और ( (0,4) ) / ( (10,0) ) and ( (0,4) ). Putting (y=0) gives (x=10), and putting (x=0) gives (y=4). Intercepts make the graph quick and clear.
Step 3
Exam Tip
(y=0) रखने पर (x=10) और (x=0) रखने पर (y=4)। अवरोधों से ग्राफ जल्दी और साफ बनता है।
At (y=0), (x=4), and at (x=0), (y=6). Intercepts help draw the line quickly.
Step 2
Why this answer is correct
The correct answer is B. ( (4,0) ) और ( (0,6) ) / ( (4,0) ) and ( (0,6) ). At (y=0), (x=4), and at (x=0), (y=6). Intercepts help draw the line quickly.
Step 3
Exam Tip
(y=0) पर (x=4) और (x=0) पर (y=6)। अवरोधों से रेखा जल्दी खींची जा सकती है।
Here (D=(-22)2-4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=11\pm\sqrt{42}\). Here (D=(-22)2-4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 3
Exam Tip
यहां (D=(-22)2-4(1)(79)=168), इसलिए \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\) है। परीक्षा में (D) को सही सरल करें।
(D=(-14)2-4(1)(13)=144), so \(x=\frac{14\pm12}{2}\) gives (1) and (13). In exams, if (D) is a perfect square, simplify quickly.
Step 2
Why this answer is correct
The correct answer is A. (x=1,13). (D=(-14)2-4(1)(13)=144), so \(x=\frac{14\pm12}{2}\) gives (1) and (13). In exams, if (D) is a perfect square, simplify quickly.
Step 3
Exam Tip
(D=(-14)2-4(1)(13)=144), इसलिए \(x=\frac{14\pm12}{2}\) से (1) और (13) मिलते हैं। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल करें।
Here (D=(-19)2-4(1)(56)=137), so \(x=\frac{19\pm\sqrt{137}}{2}\). In exams, finding (D) correctly is important.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{19\pm\sqrt{137}}{2}\). Here (D=(-19)2-4(1)(56)=137), so \(x=\frac{19\pm\sqrt{137}}{2}\). In exams, finding (D) correctly is important.
Step 3
Exam Tip
यहां (D=(-19)2-4(1)(56)=137), इसलिए \(x=\frac{19\pm\sqrt{137}}{2}\) है। परीक्षा में (D) को सही निकालना जरूरी है।
(D=(-12)2-4(1)(11)=100), so \(x=\frac{12\pm10}{2}\) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.
Step 2
Why this answer is correct
The correct answer is A. (x=1,11). (D=(-12)2-4(1)(11)=100), so \(x=\frac{12\pm10}{2}\) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.
Step 3
Exam Tip
(D=(-12)2-4(1)(11)=100), इसलिए \(x=\frac{12\pm10}{2}\) से (1) और (11) मिलते हैं। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल करें।
Here (D=(-16)2-4(1)(37)=108), so \(x=\frac{16\pm6\sqrt{3}}{2}=8\pm3\sqrt{3}\). In exams, simplify (D) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=8\pm3\sqrt{3}\). Here (D=(-16)2-4(1)(37)=108), so \(x=\frac{16\pm6\sqrt{3}}{2}=8\pm3\sqrt{3}\). In exams, simplify (D) correctly.
Step 3
Exam Tip
यहां (D=(-16)2-4(1)(37)=108), इसलिए \(x=\frac{16\pm6\sqrt{3}}{2}=8\pm3\sqrt{3}\) है। परीक्षा में (D) को सही सरल करें।
(D=(-10)2-4(1)(7)=72), so \(x=\frac{10\pm6\sqrt{2}}{2}=5\pm3\sqrt{2}\). In exams, simplify the square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=5\pm3\sqrt{2}\). (D=(-10)2-4(1)(7)=72), so \(x=\frac{10\pm6\sqrt{2}}{2}=5\pm3\sqrt{2}\). In exams, simplify the square root.
Step 3
Exam Tip
(D=(-10)2-4(1)(7)=72), इसलिए \(x=\frac{10\pm6\sqrt{2}}{2}=5\pm3\sqrt{2}\) है। परीक्षा में वर्गमूल को सरल करें।
Here (D=(-13)2-4(1)(22)=81), so \(x=\frac{13\pm9}{2}\). In exams, if (D) is a perfect square, the answer simplifies quickly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{13\pm9}{2}\). Here (D=(-13)2-4(1)(22)=81), so \(x=\frac{13\pm9}{2}\). In exams, if (D) is a perfect square, the answer simplifies quickly.
Step 3
Exam Tip
यहां (D=(-13)2-4(1)(22)=81), इसलिए \(x=\frac{13\pm9}{2}\) है। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल होता है।
(D=(-8)2-4(1)(3)=52), so \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\). In exams, simplify the square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=4\pm\sqrt{13}\). (D=(-8)2-4(1)(3)=52), so \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\). In exams, simplify the square root.
Step 3
Exam Tip
(D=(-8)2-4(1)(3)=52), इसलिए \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\) है। परीक्षा में वर्गमूल को सरल करें।
Here (D=(-10)2-4(1)(11)=56), so \(x=\frac{10\pm2\sqrt{14}}{2}=5\pm\sqrt{14}\). In exams, simplify (D) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=5\pm\sqrt{14}\). Here (D=(-10)2-4(1)(11)=56), so \(x=\frac{10\pm2\sqrt{14}}{2}=5\pm\sqrt{14}\). In exams, simplify (D) correctly.
Step 3
Exam Tip
यहां (D=(-10)2-4(1)(11)=56), इसलिए \(x=\frac{10\pm2\sqrt{14}}{2}=5\pm\sqrt{14}\) है। परीक्षा में (D) को सही सरल करें।
Here (D=(-7)2-4(1)(4)=33), so \(x=\frac{7\pm\sqrt{33}}{2}\). In exams, finding (D) correctly is important.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{7\pm\sqrt{33}}{2}\). Here (D=(-7)2-4(1)(4)=33), so \(x=\frac{7\pm\sqrt{33}}{2}\). In exams, finding (D) correctly is important.
Step 3
Exam Tip
यहां (D=(-7)2-4(1)(4)=33), इसलिए \(x=\frac{7\pm\sqrt{33}}{2}\) है। परीक्षा में (D) को सही निकालना जरूरी है।
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।
(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।
Here (D=22-4(1)(-2)=12), so \(x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}\). In exams, simplify \(\sqrt{12}=2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-1\pm\sqrt{3}\). Here (D=22-4(1)(-2)=12), so \(x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}\). In exams, simplify \(\sqrt{12}=2\sqrt{3}\).
Step 3
Exam Tip
यहां (D=22-4(1)(-2)=12), इसलिए \(x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}\) है। परीक्षा में \(\sqrt{12}=2\sqrt{3}\) सरल करें।
(D=(-10)2-4(1)(21)=16), so \(x=\frac{10\pm4}{2}\) gives (3) and (7). In exams, keep the sign of (b) correct in the formula.
Step 2
Why this answer is correct
The correct answer is A. (x=3,7). (D=(-10)2-4(1)(21)=16), so \(x=\frac{10\pm4}{2}\) gives (3) and (7). In exams, keep the sign of (b) correct in the formula.
Step 3
Exam Tip
(D=(-10)2-4(1)(21)=16), इसलिए \(x=\frac{10\pm4}{2}\) से (3) और (7) मिलते हैं। परीक्षा में सूत्र में (b) का चिन्ह सही रखें।
Here (D=1-4(1)(-1)=5), so \(x=\frac{-1\pm\sqrt{5}}{2}\). In exams, keep the sign of (c=-1) correct.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{-1\pm\sqrt{5}}{2}\). Here (D=1-4(1)(-1)=5), so \(x=\frac{-1\pm\sqrt{5}}{2}\). In exams, keep the sign of (c=-1) correct.
Step 3
Exam Tip
यहां (D=1-4(1)(-1)=5), इसलिए \(x=\frac{-1\pm\sqrt{5}}{2}\) है। परीक्षा में (c=-1) का संकेत सही रखें।
\(x^2+x-1=0\) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x-1=0\). \(x^2+x-1=0\) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.
Step 3
Exam Tip
\(x^2+x-1=0\) के सरल पूर्णांक गुणनखंड नहीं मिलते, इसलिए सूत्र विधि आसान है। परीक्षा में ऐसे मामलों में द्विघात सूत्र सुरक्षित रहता है।
Here (D=(-4)2-4(1)(-5)=36), so \(x=\frac{4\pm6}{2}\). In exams, do not forget the negative sign of (c) while using the formula.
Step 2
Why this answer is correct
The correct answer is A. (x=5,-1). Here (D=(-4)2-4(1)(-5)=36), so \(x=\frac{4\pm6}{2}\). In exams, do not forget the negative sign of (c) while using the formula.
Step 3
Exam Tip
यहां (D=(-4)2-4(1)(-5)=36), इसलिए \(x=\frac{4\pm6}{2}\) मिलता है। परीक्षा में सूत्र लगाते समय (c) का ऋण चिन्ह न भूलें।
In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 2
Why this answer is correct
The correct answer is A. ((t-2)x-2+5x+1=0), (t=2). In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 3
Exam Tip
पहले विकल्प में (t=2) रखने पर \(x^2\) का गुणांक (0) हो जाता है। तब समीकरण रैखिक बन जाता है।
\(\frac{1}{x^2}=x^{-2}\), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{x^2}+x+2=0\). \(\frac{1}{x^2}=x^{-2}\), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.
Step 3
Exam Tip
\(\frac{1}{x^2}=x^{-2}\) है, जो बहुपद रूप नहीं है। सामान्य द्विघात समीकरण में चर की ऋणात्मक घात नहीं होती।
In \(3x^2-27=0\), the \(x^2\) term is present and the (x) term is absent. An equation can be quadratic even without the (x) term.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-27=0\). In \(3x^2-27=0\), the \(x^2\) term is present and the (x) term is absent. An equation can be quadratic even without the (x) term.
Step 3
Exam Tip
\(3x^2-27=0\) में \(x^2\) पद है और (x) पद अनुपस्थित है। (x) पद न होने पर भी समीकरण द्विघात हो सकता है।
The term \(\sqrt{x}\) has a fractional power of the variable, so it is not in usual quadratic form. Quadratic form has only \(x^2\), (x), and constant terms.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{x}+x=4\). The term \(\sqrt{x}\) has a fractional power of the variable, so it is not in usual quadratic form. Quadratic form has only \(x^2\), (x), and constant terms.
Step 3
Exam Tip
\(\sqrt{x}\) में चर की भिन्न घात है, इसलिए यह सामान्य द्विघात रूप नहीं है। द्विघात रूप में केवल \(x^2\), (x) और स्थिर पद होते हैं।
In \(x^2-49=0\), the \(x^2\) term is present and the (x) term is absent. An equation can be quadratic even without an (x) term.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-49=0\). In \(x^2-49=0\), the \(x^2\) term is present and the (x) term is absent. An equation can be quadratic even without an (x) term.
Step 3
Exam Tip
\(x^2-49=0\) में \(x^2\) पद है और (x) पद अनुपस्थित है। (x) पद न होने पर भी समीकरण द्विघात हो सकता है।
In \(x+\frac{1}{x}=2\), the variable is in the denominator, so it is not directly in standard quadratic form. A quadratic polynomial form has no negative power.
Step 2
Why this answer is correct
The correct answer is C. \(x+\frac{1}{x}=2\). In \(x+\frac{1}{x}=2\), the variable is in the denominator, so it is not directly in standard quadratic form. A quadratic polynomial form has no negative power.
Step 3
Exam Tip
\(x+\frac{1}{x}=2\) में चर हर में है, इसलिए यह सीधे द्विघात मानक रूप में नहीं है। द्विघात बहुपद रूप में ऋणात्मक घात नहीं होती।
D. हाँ क्योंकि \(x^2\) का गुणांक (2) है/Yes because the coefficient of \(x^2\) is (2)
Step 1
Concept
In \(2x^2=0\), the coefficient of \(x^2\) is \(2\neq 0\). It can be quadratic even without linear and constant terms.
Step 2
Why this answer is correct
The correct answer is D. हाँ क्योंकि \(x^2\) का गुणांक (2) है / Yes because the coefficient of \(x^2\) is (2). In \(2x^2=0\), the coefficient of \(x^2\) is \(2\neq 0\). It can be quadratic even without linear and constant terms.
Step 3
Exam Tip
\(2x^2=0\) में \(x^2\) का गुणांक \(2\neq 0\) है। रैखिक और स्थिर पद न होने पर भी यह द्विघात हो सकता है।
A. हाँ क्योंकि \(x^2\) का गुणांक (1) है/Yes because the coefficient of \(x^2\) is (1)
Step 1
Concept
In \(x^2+4=0\), the coefficient of \(x^2\) is (1), so it is quadratic. Having real roots is not a condition for being quadratic.
Step 2
Why this answer is correct
The correct answer is A. हाँ क्योंकि \(x^2\) का गुणांक (1) है / Yes because the coefficient of \(x^2\) is (1). In \(x^2+4=0\), the coefficient of \(x^2\) is (1), so it is quadratic. Having real roots is not a condition for being quadratic.
Step 3
Exam Tip
\(x^2+4=0\) में \(x^2\) का गुणांक (1) है इसलिए यह द्विघात है। वास्तविक मूल होना द्विघात होने की शर्त नहीं है।
B. क्योंकि \(x^2\) का गुणांक (0) है/Because the coefficient of \(x^2\) is (0)
Step 1
Concept
Here the coefficient of \(x^2\) is (0), so the \(x^2\) term disappears. For a quadratic equation, \(a\neq 0\) is necessary.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि \(x^2\) का गुणांक (0) है / Because the coefficient of \(x^2\) is (0). Here the coefficient of \(x^2\) is (0), so the \(x^2\) term disappears. For a quadratic equation, \(a\neq 0\) is necessary.
Step 3
Exam Tip
यहाँ \(x^2\) का गुणांक (0) है इसलिए \(x^2\) पद समाप्त हो जाता है। द्विघात के लिए \(a\neq 0\) जरूरी है।
((x-7)(x-15)=x-2-22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-22x+79=0\). ((x-7)(x-15)=x-2-22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-7)(x-15)=x-2-22x+105), इसलिए \(x^2-22x+105=26\) से \(x^2-22x+79=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+6)2=49x), so \(x^2+12x+36-49x=0\), and \(x^2-37x+36=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-37x+36=0\). Cross multiplication gives ((x+6)2=49x), so \(x^2+12x+36-49x=0\), and \(x^2-37x+36=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+6)2=49x), इसलिए \(x^2+12x+36-49x=0\) और \(x^2-37x+36=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
Multiplying both sides by (7x) gives \(7+7x^2=50x\), that is \(7x^2-50x+7=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(7x^2-50x+7=0\). Multiplying both sides by (7x) gives \(7+7x^2=50x\), that is \(7x^2-50x+7=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (7x) से गुणा करने पर \(7+7x^2=50x\), यानी \(7x^2-50x+7=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
((x-6)(x-13)=x-2-19x+78), so \(x^2-19x+78=22\) gives \(x^2-19x+56=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-19x+56=0\). ((x-6)(x-13)=x-2-19x+78), so \(x^2-19x+78=22\) gives \(x^2-19x+56=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-6)(x-13)=x-2-19x+78), इसलिए \(x^2-19x+78=22\) से \(x^2-19x+56=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+5)2=36x), so \(x^2+10x+25-36x=0\), and \(x^2-26x+25=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-26x+25=0\). Cross multiplication gives ((x+5)2=36x), so \(x^2+10x+25-36x=0\), and \(x^2-26x+25=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+5)2=36x), इसलिए \(x^2+10x+25-36x=0\) और \(x^2-26x+25=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(6x^2-37x+6=0\). Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (6x) से गुणा करने पर \(6+6x^2=37x\), यानी \(6x^2-37x+6=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
((x-5)(x-11)=x-2-16x+55), so \(x^2-16x+55=18\) gives \(x^2-16x+37=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16x+37=0\). ((x-5)(x-11)=x-2-16x+55), so \(x^2-16x+55=18\) gives \(x^2-16x+37=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-5)(x-11)=x-2-16x+55), इसलिए \(x^2-16x+55=18\) से \(x^2-16x+37=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+4)2=25x), so \(x^2+8x+16-25x=0\), and \(x^2-17x+16=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-17x+16=0\). Cross multiplication gives ((x+4)2=25x), so \(x^2+8x+16-25x=0\), and \(x^2-17x+16=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+4)2=25x), इसलिए \(x^2+8x+16-25x=0\) और \(x^2-17x+16=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
Multiplying both sides by (5x) gives \(5+5x^2=26x\), that is \(5x^2-26x+5=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-26x+5=0\). Multiplying both sides by (5x) gives \(5+5x^2=26x\), that is \(5x^2-26x+5=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (5x) से गुणा करने पर \(5+5x^2=26x\), यानी \(5x^2-26x+5=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
((x-4)(x-9)=x-2-13x+36), so \(x^2-13x+36=14\) gives \(x^2-13x+22=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+22=0\). ((x-4)(x-9)=x-2-13x+36), so \(x^2-13x+36=14\) gives \(x^2-13x+22=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-4)(x-9)=x-2-13x+36), इसलिए \(x^2-13x+36=14\) से \(x^2-13x+22=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+3)2=16x), so \(x^2+6x+9-16x=0\), and \(x^2-10x+9=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+9=0\). Cross multiplication gives ((x+3)2=16x), so \(x^2+6x+9-16x=0\), and \(x^2-10x+9=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+3)2=16x), इसलिए \(x^2+6x+9-16x=0\) और \(x^2-10x+9=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-17x+4=0\). Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (4x) से गुणा करने पर \(4+4x^2=17x\), यानी \(4x^2-17x+4=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
((x-3)(x-7)=x-2-10x+21), so \(x^2-10x+21=10\) gives \(x^2-10x+11=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+11=0\). ((x-3)(x-7)=x-2-10x+21), so \(x^2-10x+21=10\) gives \(x^2-10x+11=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-3)(x-7)=x-2-10x+21), इसलिए \(x^2-10x+21=10\) से \(x^2-10x+11=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+2)2=9x), so \(x^2+4x+4-9x=0\), and \(x^2-5x+4=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-5x+4=0\). Cross multiplication gives ((x+2)2=9x), so \(x^2+4x+4-9x=0\), and \(x^2-5x+4=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+2)2=9x), इसलिए \(x^2+4x+4-9x=0\) और \(x^2-5x+4=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
Multiplying both sides by (3x) gives \(3+3x^2=10x\), that is \(3x^2-10x+3=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Multiplying both sides by (3x) gives \(3+3x^2=10x\), that is \(3x^2-10x+3=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (3x) से गुणा करने पर \(3+3x^2=10x\), यानी \(3x^2-10x+3=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
((x-2)(x-5)=x-2-7x+10), so \(x^2-7x+10=6\) gives \(x^2-7x+4=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+4=0\). ((x-2)(x-5)=x-2-7x+10), so \(x^2-7x+10=6\) gives \(x^2-7x+4=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-2)(x-5)=x-2-7x+10), इसलिए \(x^2-7x+10=6\) से \(x^2-7x+4=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Cross multiplication gives ((x+1)2=6x), so \(x^2+2x+1=6x\), and the correct form is \(x^2-4x+1=0\). In exams, cross multiply very carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-5=0\). Cross multiplication gives ((x+1)2=6x), so \(x^2+2x+1=6x\), and the correct form is \(x^2-4x+1=0\). In exams, cross multiply very carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+1)2=6x), इसलिए \(x^2+2x+1=6x\) और \(x^2-4x+1=0\) नहीं बल्कि जांच करने पर सही रूप ((x+1)2=6x) से \(x^2-4x+1=0\) बनता है। परीक्षा में क्रॉस गुणा बहुत सावधानी से करें।
Multiplying both sides by (2x) gives \(2+2x^2=5x\), that is \(2x^2-5x+2=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-5x+2=0\). Multiplying both sides by (2x) gives \(2+2x^2=5x\), that is \(2x^2-5x+2=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (2x) से गुणा करने पर \(2+2x^2=5x\), यानी \(2x^2-5x+2=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
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।
If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+21=0\). If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (3) और (7) हों तो ((x-3)(x-7)=0), यानी \(x^2-10x+21=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (2) और (5) हों तो समीकरण ((x-2)(x-5)=0) यानी \(x^2-7x+10=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
In the quadratic formula, the part inside the square root is \(b^2-4ac\). In exams, it is also called the discriminant (D).
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4ac\). In the quadratic formula, the part inside the square root is \(b^2-4ac\). In exams, it is also called the discriminant (D).
Step 3
Exam Tip
द्विघात सूत्र में वर्गमूल के अंदर \(b^2-4ac\) होता है। परीक्षा में इसे विविक्तकर (D) भी कहते हैं।
A. वास्तविक मूल नहीं होंगे/There will be no real roots
Step 1
Concept
When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं होंगे / There will be no real roots. When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 3
Exam Tip
(D<0) होने पर वास्तविक वर्गमूल नहीं मिलता। परीक्षा में ऋणात्मक विविक्तकर का अर्थ याद रखें।
A. दो अलग वास्तविक मूल मिलेंगे/Two distinct real roots will be obtained
Step 1
Concept
(D=36>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. दो अलग वास्तविक मूल मिलेंगे / Two distinct real roots will be obtained. (D=36>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
(D=36>0), इसलिए दो अलग वास्तविक मूल मिलते हैं। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
A. वास्तविक मूल नहीं होंगे/There will be no real roots
Step 1
Concept
When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं होंगे / There will be no real roots. When (D<0), no real square root is obtained. In exams, remember the meaning of a negative discriminant.
Step 3
Exam Tip
(D<0) होने पर वास्तविक वर्गमूल नहीं मिलता। परीक्षा में ऋणात्मक विविक्तकर का अर्थ याद रखें।
