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।
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 \(2x^2+7x=0\), the \(x^2\) term is present and the constant term is absent. An equation can be quadratic even without a constant term.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+7x=0\). In \(2x^2+7x=0\), the \(x^2\) term is present and the constant term is absent. An equation can be quadratic even without a constant term.
Step 3
Exam Tip
\(2x^2+7x=0\) में \(x^2\) पद है और स्थिर पद अनुपस्थित है। स्थिर पद न होने पर भी समीकरण द्विघात हो सकता 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।
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 \(x+\frac{1}{x}=2\), 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।
Here (ac=900) and (-36+(-25)=-61), so the correct split is (-36x-25x). In exams, even for large (ac), match both sum and product.
Step 2
Why this answer is correct
The correct answer is A. \(30x^2-36x-25x+30=0\). Here (ac=900) and (-36+(-25)=-61), so the correct split is (-36x-25x). In exams, even for large (ac), match both sum and product.
Step 3
Exam Tip
यहां (ac=900) और (-36+(-25)=-61), इसलिए सही विभाजन (-36x-25x) है। परीक्षा में बड़ा (ac) हो तो भी योग और गुणनफल दोनों मिलाएं।
Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.
Step 2
Why this answer is correct
The correct answer is A. \(24x^2-30x-20x+25=0\). Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.
Step 3
Exam Tip
यहां (ac=600) और (-30+(-20)=-50), इसलिए सही विभाजन (-30x-20x) है। परीक्षा में बड़ा (ac) हो तो भी योग और गुणनफल दोनों मिलाएं।
Here (ac=420) and (-28+(-15)=-43), so the correct split is (-28x-15x). In exams, even when (ac) is large, match both sum and product.
Step 2
Why this answer is correct
The correct answer is A. \(20x^2-28x-15x+21=0\). Here (ac=420) and (-28+(-15)=-43), so the correct split is (-28x-15x). In exams, even when (ac) is large, match both sum and product.
Step 3
Exam Tip
यहां (ac=420) और (-28+(-15)=-43), इसलिए सही विभाजन (-28x-15x) है। परीक्षा में (ac) बड़ा हो तो भी योग और गुणनफल दोनों मिलाएं।
Here (ac=180) and (-15+(-12)=-27), so the correct split is (-15x-12x). In exams, match both sum (b) and product (ac).
Step 2
Why this answer is correct
The correct answer is A. \(18x^2-15x-12x+10=0\). Here (ac=180) and (-15+(-12)=-27), so the correct split is (-15x-12x). In exams, match both sum (b) and product (ac).
Step 3
Exam Tip
यहां (ac=180) और (-15+(-12)=-27), इसलिए सही विभाजन (-15x-12x) है। परीक्षा में योग (b) और गुणनफल (ac) दोनों मिलाएं।
Here (ac=60) and (10+6=16), so (16x) is split as (10x+6x). In exams, check both sum (b) and product (ac).
Step 2
Why this answer is correct
The correct answer is A. \(15x^2+10x+6x+4=0\). Here (ac=60) and (10+6=16), so (16x) is split as (10x+6x). In exams, check both sum (b) and product (ac).
Step 3
Exam Tip
यहां (ac=60) और (10+6=16), इसलिए (16x) को (10x+6x) में तोड़ते हैं। परीक्षा में योग (b) और गुणनफल (ac) दोनों जांचें।
Here (ac=72) and (-9+(-8)=-17), so the correct split is (-9x-8x). In exams, check both sum (b) and product (ac).
Step 2
Why this answer is correct
The correct answer is A. \(12x^2-9x-8x+6=0\). Here (ac=72) and (-9+(-8)=-17), so the correct split is (-9x-8x). In exams, check both sum (b) and product (ac).
Step 3
Exam Tip
यहां (ac=72) और (-9+(-8)=-17), इसलिए सही विभाजन (-9x-8x) है। परीक्षा में योग (b) और गुणनफल (ac) दोनों जांचें।
(-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।
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।
(-8+(-10)=-18) and ((-8)(-10)=80), 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 (-10) / (-8) and (-10). (-8+(-10)=-18) and ((-8)(-10)=80), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.
Step 3
Exam Tip
(-8+(-10)=-18) और ((-8)(-10)=80), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक होती हैं।
Here (ac=-30) and (-15+2=-13), so the middle term is (-15x+2x). In exams, check the sign of (ac) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-15x+2x-6=0\). Here (ac=-30) and (-15+2=-13), so the middle term is (-15x+2x). In exams, check the sign of (ac) carefully.
Step 3
Exam Tip
यहां (ac=-30) और (-15+2=-13), इसलिए मध्य पद (-15x+2x) होगा। परीक्षा में (ac) का संकेत ध्यान से देखें।
(-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.
Step 2
Why this answer is correct
The correct answer is A. (-5) और (-9) / (-5) and (-9). (-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.
Step 3
Exam Tip
(-5+(-9)=-14) और ((-5)(-9)=45), इसलिए यह जोड़ी सही है। परीक्षा में (c) धनात्मक और (b) ऋणात्मक हो तो दोनों संख्याएं ऋणात्मक हो सकती हैं।
((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-9x-2x+6=0\). ((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).
Step 3
Exam Tip
((-9)+(-2)=-11) और ((-9)(-2)=18=ac), इसलिए (-9x-2x) सही है। परीक्षा में (ac) देखकर मध्य पद तोड़ें।
(12+1=13) and \(12\times1=12=ac\), so (13x) is split as (12x+x). In exams, keep both sum and product correct.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2+12x+x+3=0\). (12+1=13) and \(12\times1=12=ac\), so (13x) is split as (12x+x). In exams, keep both sum and product correct.
Step 3
Exam Tip
(12+1=13) और \(12\times1=12=ac\), इसलिए (13x) को (12x+x) में तोड़ते हैं। परीक्षा में योग और गुणनफल दोनों सही रखें।
(8+(-5)=3) and \(8\times(-5)=-40\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 2
Why this answer is correct
The correct answer is A. (8) और (-5) / (8) and (-5). (8+(-5)=3) and \(8\times(-5)=-40\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 3
Exam Tip
(8+(-5)=3) और \(8\times(-5)=-40\), इसलिए यह जोड़ी सही है। परीक्षा में योग (b) और गुणनफल (c) से मिलाएं।
((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-5x+5=0\). ((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.
Step 3
Exam Tip
((-2)+(-5)=-7) और ((-2)(-5)=10=ac), इसलिए (-2x-5x) सही है। परीक्षा में मध्य पद तोड़ते समय (ac) देखना जरूरी है।
(6+(-4)=2) and \(6\times(-4)=-24\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 2
Why this answer is correct
The correct answer is A. (6) और (-4) / (6) and (-4). (6+(-4)=2) and \(6\times(-4)=-24\), so this pair is correct. In exams, match the sum with (b) and product with (c).
Step 3
Exam Tip
(6+(-4)=2) और \(6\times(-4)=-24\), इसलिए यह जोड़ी सही है। परीक्षा में योग (b) और गुणनफल (c) से मिलाएं।
((-2)+(-3)=-5) and ((-2)(-3)=6=ac), so the correct split is (-2x-3x). In exams, the product of the two numbers must be (ac).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-3x+3=0\). ((-2)+(-3)=-5) and ((-2)(-3)=6=ac), so the correct split is (-2x-3x). In exams, the product of the two numbers must be (ac).
Step 3
Exam Tip
((-2)+(-3)=-5) और ((-2)(-3)=6=ac), इसलिए सही विभाजन (-2x-3x) है। परीक्षा में दोनों संख्याओं का गुणनफल (ac) होना चाहिए।
There is no (x) term, so the linear term is absent. The coefficient of a missing term is considered (0).
Step 2
Why this answer is correct
The correct answer is C. रैखिक पद / Linear term. There is no (x) term, so the linear term is absent. The coefficient of a missing term is considered (0).
Step 3
Exam Tip
इसमें (x) वाला पद नहीं है इसलिए रैखिक पद अनुपस्थित है। अनुपस्थित पद का गुणांक (0) माना जाता 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।
