If the roots of (4x^2-(p-1)x+9=0) are (\frac{3}{2}) and (\frac{3}{2}), what is (p)?
Correct answer: A
The sum of roots is (3), and (\frac{p-1}{4}=3), so (p=13). In exams, use (-\frac{b}{a}) for the sum of roots.
Mathematics
Methods of Solving Quadratic Equations
द्विघात समीकरणों को हल करने की विधियाँ
In this Class 10 Mathematics topic from the chapter Quadratic Equations, students learn how to find the values of an unknown variable that satisfy a quadratic equation. They practise solving equations by factorisation, completing the square, and using the quadratic formula, while learning when each method is useful. The topic also develops skills in identifying coefficients, calculating the discriminant, checking solutions, and interpreting whether an equation has two, one, or no real roots.
Hard · Level 5 · 25 questions
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Hard · Level 5
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- (p=13)
- (p=11)
- (p=9)
- (p=7)
Hard · Level 5
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- \(\left(x-\frac{11}{7}\right)^2=\frac{72}{49}\)
- \(\left(x+\frac{11}{7}\right)^2=\frac{72}{49}\)
- \(\left(x-\frac{22}{7}\right)^2=1\)
- \(\left(x-\frac{11}{7}\right)^2=\frac{121}{49}\)
Hard · Level 5
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- \(x=\frac{11\pm6\sqrt{2}}{7}\)
- \(x=\frac{-11\pm6\sqrt{2}}{7}\)
- \(x=\frac{22\pm6\sqrt{2}}{7}\)
- \(x=\frac{11\pm\sqrt{2}}{7}\)
Hard · Level 5
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- (m=81)
- (m=18)
- (m=36)
- (m=324)
Hard · Level 5
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- (x=\frac{3}{2})
- (x=\frac{4}{5})
- (x=\frac{4}{3})
- (x=2)
Hard · Level 5
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- (x=\frac{3}{2})
- (x=\frac{1}{2})
- (x=\frac{4}{3})
- (x=2)
Hard · Level 5
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- (9)
- (6)
- (15)
- (-9)
Hard · Level 5
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- (54)
- (90)
- (9)
- (15)
Hard · Level 5
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- Correct
- Incorrect because the second root is (\frac{3}{2})
- Incorrect because the first root is (-\frac{1}{2})
- Incorrect because both roots are positive
Hard · Level 5
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- ((x+\sqrt{7})^2=0)
- ((x-\sqrt{7})^2=0)
- ((x+7)^2=0)
- ((x+2\sqrt{7})^2=0)
Hard · Level 5
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- (x=-\sqrt{7})
- (x=\sqrt{7})
- (x=-7)
- (x=7)
Hard · Level 5
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- (x=u,v)
- (x=-u,-v)
- (x=u+v,uv)
- (x=u-v,v-u)
Hard · Level 5
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- (x=m+n,m-n)
- (x=-m+n,-m-n)
- (x=n+m,n-m)
- (x=m^2,n^2)
Hard · Level 5
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- (x=a,b)
- (x=-a,-b)
- (x=16a,16b)
- (x=a+b,ab)
Hard · Level 5
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- (y^2-13y+36=0)
- (y^2+13y+36=0)
- (y^2-13x+36=0)
- (y-13y+36=0)
Hard · Level 5
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- (x=\pm2,\pm3)
- (x=2,3)
- (x=-2,-3)
- (x=\pm6,\pm13)
Hard · Level 5
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- (4x^2-17x+4=0)
- (4x^2+17x+4=0)
- (x^2-17x+4=0)
- (4x^2-17=0)
Hard · Level 5
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- (x=4,\frac{1}{4})
- (x=-4,-\frac{1}{4})
- (x=17,4)
- (x=\frac{17}{4},1)
Hard · Level 5
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- (x^2-10x+9=0)
- (x^2+6x-7=0)
- (x^2-16x+9=0)
- (x^2+10x+9=0)
Hard · Level 5
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- (x=1,9)
- (x=-1,-9)
- (x=3,16)
- (x=10,9)
Hard · Level 5
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- (x=4\pm\sqrt{13})
- (x=-4\pm\sqrt{13})
- (x=8\pm\sqrt{13})
- (x=4\pm13)
Hard · Level 5
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- (35)
- (-35)
- (7)
- (100)
Hard · Level 5
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- (3)
- (6)
- ( \frac{1}{12}) / (\frac{1}{12})
- ( \frac{13}{2}) / (\frac{13}{2})
Hard · Level 5
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- (149)
- (289)
- (70)
- (219)
Hard · Level 5
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- ( \frac{12}{35}) / (\frac{12}{35})
- ( \frac{35}{12}) / (\frac{35}{12})
- (12)
- (35)
Which middle step is correct while solving \(7x^2-22x+7=0\) by completing the square?
Correct answer: A
First \(x^2-\frac{22}{7}x+1=0\) is obtained, then \(\left(x-\frac{11}{7}\right)^2=\frac{72}{49}\). In exams, divide by (a) first when \(a\neq1\).
What will be the roots of \(7x^2-22x+7=0\)?
Correct answer: A
Since \(\left(x-\frac{11}{7}\right)^2=\frac{72}{49}\), \(x=\frac{11\pm6\sqrt{2}}{7}\). In exams, simplify \(\sqrt{72}=6\sqrt{2}\).
If (x^2-18x+m=0) has real and equal roots, what is the value of (m)?
Correct answer: A
The direct answer is option A, m=81. For a quadratic equation ax^2+bx+c=0, the discriminant is D=b^2-4ac. Real and equal roots occur exactly when D=0, because the square-root part of the quadratic formula must become zero. Here a=1, b=-18, and c=m. Therefore D=(-18)^2-4(1)(m)=324-4m. Set it equal to zero: 324-4m=0, so 4m=324 and m=81. Thus option A is correct. Option B, m=18, gives D=324-72=252, so the roots are real but unequal. Option C, m=36, gives D=324-144=180, also unequal roots. Option D, m=324, gives D=324-1296=-972, so the roots are not real. A quick alternative is to use the equal-root form: x^2-18x+81=(x-9)^2. Memory cue: equal roots always mean discriminant zero.
Which root is common to (5x^2-16x+12=0) and (6x^2-17x+12=0)?
Correct answer: A
The first equation has roots (2,\frac{6}{5}), and the second has roots (\frac{3}{2},\frac{4}{3}), so there is no common root among the given values. In exams, solve both equations before comparing.
Which root is common to (4x^2-12x+5=0) and (6x^2-17x+12=0)?
Correct answer: A
The first equation has roots (\frac{1}{2},\frac{5}{2}), and the second has roots (\frac{3}{2},\frac{4}{3}), so none of the listed values is common. In exams, solve both equations correctly before comparing.
If one root of (x^2-15x+q=0) is (6), what will be the other root?
Correct answer: A
The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
If one root of (x^2-15x+q=0) is (6), what is the value of (q)?
Correct answer: A
The other root is (9), so (q=6\times9=54). In exams, when (a=1), (c) equals the product of roots.
A student wrote (x=\frac{1}{2},-\frac{3}{2}) for (4x^2+4x-3=0). How is this answer?
Correct answer: A
(4x^2+4x-3=(2x-1)(2x+3)), so (x=\frac{1}{2},-\frac{3}{2}) is correct. In exams, change signs carefully from factors.
In which form can (x^2+2\sqrt{7}x+7=0) be written?
Correct answer: A
Since (7=(\sqrt{7})^2) and the middle term is (2\sqrt{7}x), it is ((x+\sqrt{7})^2). In exams, identify perfect squares even with irrational coefficients.
What is the root of (x^2+2\sqrt{7}x+7=0)?
Correct answer: A
The direct answer is option A: the repeated root is \(x=-\sqrt{7}\). Since \((\sqrt{7})^2=7\) and the middle term is \(2x\sqrt{7}\), the quadratic becomes \(x^2+2\sqrt{7}x+7=(x+\sqrt{7})^2\). Hence \((x+\\sqrt{7})^2=0\), so \(x+ \sqrt{7}=0\) and \(x=-\sqrt{7}\). Option A is correct. Option B has the wrong sign; substituting \(\sqrt{7}\) does not make the expression zero. Option C, \(-7\), treats the square root as the number itself and is not a solution. Option D, 7, has both the wrong sign and wrong magnitude. Because the square is zero, both algebraic roots are the same, so this is a repeated root. Memory cue: match \(x^2+2ax+a^2\) with \((x+a)^2\), then change the sign of a.
What are the roots of (x^2-(u+v)x+uv=0)?
Correct answer: A
The equation is equivalent to ((x-u)(x-v)=0), so the roots are (u) and (v). In exams, apply the same rule to symbolic factors.
If (x^2-2mx+m^2-n^2=0), what are its roots?
Correct answer: A
It is ((x-m)^2-n^2=0), so (x-m=\pm n) and (x=m\pm n). In exams, recognize the difference of squares.
After simplifying (16x^2-16(a+b)x+16ab=0), what will be the roots?
Correct answer: A
Dividing the whole equation by (16) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.
If (y=x^2) is put in (x^4-13x^2+36=0), which equation is obtained?
Correct answer: A
Since (x^4=(x^2)^2=y^2), the new equation is (y^2-13y+36=0). In exams, substitution simplifies a difficult form.
What are the real solutions of (x^4-13x^2+36=0)?
Correct answer: A
From (y^2-13y+36=0), (y=4,9), so (x^2=4,9) and (x=\pm2,\pm3). In exams, do not forget to return to (x).
For (\frac{1}{x}+x=\frac{17}{4}), (x\neq0), what quadratic form is obtained?
Correct answer: A
Multiplying both sides by (4x) gives (4+4x^2=17x), that is (4x^2-17x+4=0). In exams, remember the condition (x\neq0).
What are the solutions of (\frac{1}{x}+x=\frac{17}{4}), (x\neq0)?
Correct answer: A
(4x^2-17x+4=(4x-1)(x-4)), so (x=\frac{1}{4}) and (4). In exams, check whether obtained roots are valid in the original equation.
What is the correct quadratic form of (\frac{x+3}{x}=\frac{16}{x+3}), (x\neq0,-3)?
Correct answer: A
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.
What are the solutions of (\frac{x+3}{x}=\frac{16}{x+3}), (x\neq0,-3)?
Correct answer: A
(x^2-10x+9=(x-1)(x-9)), so (x=1) and (x=9). In exams, check solutions against excluded denominator values.
What are the roots of (x^2-8x+3=0) by quadratic formula?
Correct answer: A
(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.
If the sum of roots of (5x^2+px+20=0) is (-7), what is (p)?
Correct answer: A
The sum of roots is (-\frac{p}{5}), so (-\frac{p}{5}=-7) gives (p=35). In exams, remember the sum formula (-\frac{b}{a}).
If the product of roots of (6x^2-13x+p=0) is (\frac{1}{2}), what is (p)?
Correct answer: A
The product of roots is (\frac{p}{6}), so (\frac{p}{6}=\frac{1}{2}) gives (p=3). In exams, use the product formula (\frac{c}{a}).
If the roots of (x^2-17x+70=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?
Correct answer: A
(\alpha+\beta=17) and (\alpha\beta=70), so (\alpha^2+\beta^2=17^2-2(70)=149). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).
If the roots of (x^2-12x+35=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?
Correct answer: A
(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}). In exams, first write sum and product in reciprocal questions.
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