CBSE Class 10 Mathematics Quadratic Equations VBQs Set I

Quadratic Equations

Question. If \( y = 1 \) is a common root of the equations \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \), then \( ab \) equals :*
(a) 3
(b) \( -\frac{7}{2} \)
(c) 6
(d) – 3
Answer: (a) 3
Explanation : The given equations are \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \). Substituting \( y = 1 \) in them we have \( a(1)^2 + a(1) + 3 = 0 \) and \( (1)^2 + (1) + b = 0 \). \( \Rightarrow 2a + 3 = 0 \) and \( 2 + b = 0 \). \( \Rightarrow a = -\frac{3}{2} \) and \( b = -2 \). Thus, \( ab = \left(-\frac{3}{2}\right) (-2) = 3 \)

Question. The roots of the equation \( x^2 - 3x - 9 = 0 \) are :*
(a) real and unequal
(b) real and equal
(c) roots are not equal
(d) imaginary roots
Answer: (a) real and unequal
Explanation : \( D = b^2 - 4ac = (-3)^2 - 4(1)(-9) = 9 + 36 = 45 > 0 \). Thus, the roots are real and distinct.

Question. The value of \( x \) in \( x - \frac{18}{x} = 6 \) is :*
(a) real and unequal
(b) real and equal
(c) roots are imaginary
(d) roots are not equal
Answer: (a) real and unequal
Explanation : Given, \( x - \frac{18}{x} = 6 \) or \( x^2 - 6x - 18 = 0 \). Thus, \( D = b^2 - 4ac = (-6)^2 - 4(-18) = 36 + 72 = 108 > 0 \). Thus, the roots are real and distinct.

Question. The roots of the quadratic equation \( x^2 - 0.04 = 0 \) are :*
(a) ± 0.2
(b) ± 0.02
(c) 0.4
(d) 2
Answer: (a) ± 0.2
Explanation : Given : \( x^2 - 0.04 = 0 \Rightarrow x^2 - (0.2)^2 = 0 \Rightarrow (x + 0.2) (x - 0.2) = 0 \Rightarrow x = -0.2, 0.2 \)

Question. Values of \( k \) for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots is:
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8
Answer: (d) 0, 8
Explanation : For equal roots \( D = 0 \Rightarrow b^2 - 4ac = 0 \Rightarrow k^2 - 4 \times 2 \times k = 0 \Rightarrow k^2 - 8k = 0 \Rightarrow k(k - 8) = 0 \Rightarrow k = 0, 8 \)

Question. The roots of the equation \( x^2 - x - 6 \) are:
(a) real and equal
(b) real and unequal
(c) unreal
(d) none of the options
Answer: (b) real and unequal
Explanation : sum of root = -3 + 3 = 0 product of roots = -3 × 3 = - 9 quadratic equation = \( x^2 - 0 \times x - 9 = 0 \Rightarrow x^2 = 9 = 0 \)

Question. The quadratic equation whose roots are 3 and – 3 is:
(a) \( x^2 + 9 = 0 \)
(b) \( x^2 - 9 = 0 \)
(c) \( 2x^2 - 3 = 0 \)
(d) \( x^2 + 2x + 2 = 0 \)
Answer: (b) \( x^2 - 9 = 0 \)
Explanation : \( D = b^2 - 4ac \Rightarrow (-1)^2 - 4(-6)(1) = 1 + 24 = 25 \Rightarrow \) so roots are real and distinct (or equal)

Question. If \( \alpha \) and \( \beta \) are the zeroes of \( x^2 + 5x + 8 \) then the value of \( \alpha + \beta \) is:
(a) 5
(b) 8
(c) – 5
(d) – 8
Answer: (c) – 5
Explanation : Sum of roots \( = \alpha + \beta = -\left(\frac{5}{1}\right) = -5 \)

Question. The discriminant of the quadratic equation \( 3x^2 - 4x - 2 = 0 \) is equal to:
(a) 40
(b) 20
(c) 24
(d) 48
Answer: (a) 40
Explanation : \( D = b^2 - 4ac = (-4)^2 - 4 \times 3 \times (-2) = 16 + 24 = 40 \).

Question. If \( ax^2 + bx + c = 0 \) has equal roots, then \( c = \)
(a) \( \frac{-b}{2a} \)
(b) \( \frac{b^2}{4a} \)
(c) \( \frac{-b^2}{4a} \)
(d) \( \frac{b^2}{4a} \)
Answer: (d) \( \frac{b^2}{4a} \)
Explanation : If \( ax^2 + bx + c = 0 \) has equal roots, then \( D = b^2 - 4ac = 0 \Rightarrow c = \frac{b^2}{4a} \)

Question. If one root of the equation \( 2x^2 + 3x + c = 0 \) is 0.5, then what is the value of \( c \) ?
(a) – 1
(b) – 2
(c) – 3
(d) – 4
Answer: (b) – 2

Answer: Explanation : \( 2x^2 + 3x + c = 0 \) is 0.5, \( 2(0.5)^2 + 3(0.5) + c = 0 \Rightarrow 0.5 + 1.5 + c = 0 \Rightarrow c = -2 \)

Question. The equation whose roots are twice the roots of the equation \( x^2 - 2x + 4 = 0 \) is :
(a) \( x^2 - 2x + 4 = 0 \)
(b) \( x^2 - 2x + 16 = 0 \)
(c) \( x^2 - 4x + 8 = 0 \)
(d) \( x^2 - 4x + 16 = 0 \)
Answer: (d) \( x^2 - 4x + 16 = 0 \)
Explanation : \( \alpha + \beta = 2 \) and \( \alpha\beta = 4 \). On taking \( \alpha \rightarrow 2\alpha \), \( \beta \rightarrow 2\beta \). \( 2\alpha + 2\beta = 4 \) and \( 2\alpha.2\beta = 4 \times 4 = + 16 \). \( x^2 - 4x + 16 = 0 \)

Question. The difference in the roots of the equation \( 2x^2 - 11x + 5 = 0 \) is :
(a) 4.5
(b) 4
(c) 3.5
(d) 3
Answer: (a) 4.5
Explanation : Let \( \alpha \) and \( \beta \) be the root of this quadratic equation \( 2x^2 - 11x + 5 = 0 \). \( \alpha + \beta = (11/2) \), \( \alpha\beta = (5/2) \). We know that, \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4 \alpha\beta = (\frac{11}{2})^2 - 4(\frac{5}{2}) = \frac{121}{4} - 10 = \frac{81}{4} \). Difference of roots \( = (\alpha - \beta) = 4.5 \)

Question. If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + px + q = 0 \), then what is value of \( \alpha^2 + \beta^2 \)?
(a) \( p^2 - 2q \)
(b) \( q^2 - 2p \)
(c) \( p^2 + 2q \)
(d) \( q^2 - p \)
Answer: (a) \( p^2 - 2q \)
Explanation : We know that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + px + q = 0 \), \( \alpha + \beta = -p \) and \( \alpha\beta = q \). Now, \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-p)^2 - 2(q) = p^2 - 2q \)

Question. In the quadratic equation \( x^2 + ax + b = 0 \), \( a \) and \( b \) can take any value from the set {1, 2, 3, 4}. How many pairs of values of \( a \) and \( b \) are possible in order that the quadratic equation has real roots ?
(a) 6
(b) 7
(c) 8
(d) 16
Answer: (b) 7
Explanation : For real roots, \( B^2 - 4AC \geq 0 \). So, by equation, \( a^2 - 4b \geq 0 \Rightarrow a^2 \geq 4b \). When \( b = 1 \), then \( a^2 \geq 4 \Rightarrow a^2 - 4 \geq 0 \Rightarrow a = 2, 3, 4 \). When \( b = 2 \), then \( a^2 - 8 > 0 \Rightarrow b = 2 \) and \( a = 3, 4 \). When \( b = 3 \), then \( a^2 - 12 > 0 \Rightarrow b = 3 \) and \( a = 4 \). When \( b = 4 \), then \( a^2 - 16 \geq 0 \Rightarrow a = 4 \) and \( b = 4 \). Hence, 7 pairs of values of \( a \) and \( b \) are possible

Question. For which value of \( k \) does the pair of equation \( x^2 - y^2 = 0 \) and \( (x - k)^2 + y^2 = 1 \) yield a unique positive solution of \( x \) ?
(a) 2
(b) 0
(c) \( \sqrt{2} \)
(d) \( -\sqrt{2} \)
Answer: (c) \( \sqrt{2} \)
Explanation : \( x^2 - y^2 = 0 \) ...(i) and \( (x - k)^2 + y^2 = 1 \Rightarrow x^2 + k^2 - 2kx + y^2 - 1 = 0 \) ...(ii). From equations (i) and (ii): \( 2x^2 - 2kx + k^2 - 1 = 0 \). For unique solution \( b^2 - 4ac = 0 \) must satisfy \( (- 2k)^2 - 4 \times 2 \times (k^2 - 1) = 0 \Rightarrow 4k^2 = 8 \Rightarrow k = \sqrt{2} \)

Question. If the root of the equation \( Ax^2 - Bx + C = 0 \) are –1 and 1, then which one of the following is correct ?
(a) A and C are both zero
(b) A and B are both positive
(c) A and C are both negative
(d) A and C are of opposite sign
Answer: (d) A and C are of opposite sign
Explanation : \( Ax^2 - Bx + C = 0 \). Since the given roots are –1 and 1. Sum of roots = \( -1 + 1 = 0 \). Product of roots = \( 1 \times (-1) = -1 \). \( x^2 - (\text{sum of root})x + \text{product of roots} = 0 \Rightarrow x^2 + (\frac{B}{A})x + (\frac{C}{A}) = 0 \). \( \frac{C}{A} = \text{Product of roots} = - 1 \Rightarrow C = -A \)

Question. If the roots of the equation \( (a^2 - bc)x^2 + 2(b^2 - ac)x + (c^2 - ab) = 0 \) are equal, where \( b \neq 0 \), then which one of the following is correct ?
(a) \( a + b + c = abc \)
(b) \( a^2 + b^2 + c^2 = 0 \)
(c) \( a^3 + b^3 + c^3 = 3a \)
(d) \( a^3 + b^3 + c^3 = 3abc \)
Answer: (c) \( a^3 + b^3 + c^3 = 3a \)
Explanation : \( (a^2 - bc)x^2 + 2(b^2 - ac)x + (c^2 - ab) = 0 \). The given roots are equal, and then \( D \) must be zero. \( [2(b^2 - ac)]^2 - 4(a^2 - bc)(c^2 - ab) = 0 \Rightarrow 4b^4 - 12ab^2c + 4bc^3 + 4a^3b = 0 \Rightarrow a^3 + b^3 + c^3 = 3a \)

Question. If one root of \( (a^2 - 5a + 3) x^2 + (3a - 1) x + 2 = 0 \) is twice the other, then what is the value of \( a \) ?
(a) \( \frac{2}{3} \)
(b) \( -\frac{2}{3} \)
(c) \( \frac{1}{3} \)
(d) \( -\frac{1}{3} \)
Answer: (a) \( \frac{2}{3} \)
Explanation : Let \( \alpha \) and \( 2\alpha \) be the roots of the given equation. \( \alpha + 2\alpha = \frac{-(3a-1)}{a^2-5a+3} \Rightarrow 3\alpha = \frac{-(3a-1)}{a^2-5a+3} \Rightarrow \alpha = \frac{-(3a-1)}{3(a^2-5a+3)} \) ...(i). Also \( (\alpha)(2\alpha) = \frac{2}{a^2-5a+3} \Rightarrow \alpha^2 = \frac{1}{a^2-5a+3} \) ...(ii). From eqns. (i) and (ii), we get \( \left[\frac{-(3a-1)}{9(a^2-5a+3)}\right]^2 = \frac{1}{a^2-5a+3} \Rightarrow \frac{-(3a-1)^2}{9(a^2-5a+3)^2} = \frac{1}{a^2-5a+3} \Rightarrow (3a - 1)^2 = 9(a^2 - 5a + 3) \Rightarrow 9a^2 - 6a + 1 = 9a^2 - 45a + 27 \Rightarrow 39a = 26 \Rightarrow a = \frac{26}{39} = \frac{2}{3} \)

Question. If 3 is a root of the equation \( kx^2 - kx - 3 = 0 \), then the value of \( k \) is ........... .
(a) 3
(b) 1/2
(c) 4
(d) 2
Answer: (b) 1/2
 Explanation : If 3 is the root of \( kx^2 - kx - 3 = 0 \) then it satisfy it i.e., \( 9k - 3k - 3 = 0 \Rightarrow 6k - 3 = 0 \Rightarrow 6k = 3 \Rightarrow k = \frac{3}{6} \Rightarrow k = \frac{1}{2} \)

Question. If the zeros of the quadratic equation \( x^2 + (a + 1) x + b + 1 = 0 \) are 2 and – 3, then the values of \( a \) and \( b \), are ........... and ........... respectively.
(a) 0, 7
(b) 0, – 7
(c) 2, 3
(d) none of the options
Answer: (b) 0, – 7

Question. The roots of the equation \( \sqrt{2x+9} + x = 13 \) are ....... and ............. .
(a) – 8, – 20
(b) – 8, 20
(c) 8, 20
(d) 8, – 20
Answer: (c) 8, 20

Question. The nature of roots of the equation \( 2x^2 + \sqrt{5}x - 1 = 0 \) is ........... .
(a) real and equal
(b) imaginary and equal
(c) imaginary and unequal
(d) real and unequal
Answer: (d) real and unequal
Explanation : \( 2x^2 + \sqrt{5}x - 1 = 0 \). Here, \( b = \sqrt{5}, a = 2, c = -1 \). \( b^2 - 4ac \Rightarrow (\sqrt{5})^2 - 4 \times 2 \times (-1) = 5 - (-8) = 13 > 0 \). Hence, roots are real and unequal.

Question. Constant term that must be added to the equation \( 2x^2 - 5x + 3 = 0 \) to solve it by the method of completing the square is ........... .
(a) \( \frac{2.5}{16} \)
(b) \( \frac{25}{16} \)
(c) \( \frac{25}{4} \)
(d) \( \frac{5}{16} \)
Answer: (b) \( \frac{25}{16} \)

Question. Which of the following is not a quadratic equation?
(a) \( 2(x + 1)^2 = 4x^2 - 2x + 1 \)
(b) \( 2x - x^2 = x^2 + 5 \)
(c) \( (\sqrt{2}x + \sqrt{3}x)^2 + x^2 = 3x^2 - 5x \)
(d) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^2 \)
Answer: (d) (x^2 + 2x)^2 = x^4 + 3 + 4x^2
Explanation : Given that, \( (x^2 + 2x)^2 = x^4 + 3 + 4x^2 \) \( \Rightarrow x^4 + 4x^2 + 4x^3 = x^4 + 3 + 4x^2 \Rightarrow 4x^3 - 3 = 0 \) which is not of the form \( ax^2 + bx + c, a \neq 0 \). Thus, the equation is not quadratic. This is a cubic equation.

Question. Which of the following equation has 2 as a root?
(a) \( x^2 - 4x + 5 = 0 \)
(b) \( x^2 + 3x - 12 = 0 \)
(c) \( 2x^2 - 7x + 6 = 0 \)
(d) \( 3x^2 - 6x - 2 = 0 \)
Answer: (c) 2x^2 - 7x + 6 = 0
Explanation : Substituting \( x = 2 \) in \( 2x^2 - 7x + 6 \), we get \( 2(2)^2 - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 14 - 14 = 0 \). So, \( x = 2 \) is root of the equation \( 2x^2 - 7x + 6 = 0 \).

Question. If \( \frac{1}{2} \) is a root of the equation \( x^2 + kx - \frac{5}{4} = 0 \), then the value of \( k \) is :
(a) 2
(b) - 2
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{2} \)
Answer: (a) 2
Explanation : Since, \( \frac{1}{2} \) is a root of the quadratic equation \( x^2 + kx - \frac{5}{4} = 0 \). Then, \( (\frac{1}{2})^2 + k(\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0 \Rightarrow \frac{1 + 2k - 5}{4} = 0 \Rightarrow 2k - 4 = 0 \Rightarrow 2k = 4 \Rightarrow k = 2 \).

Question. Which one of the following is a quadratic equation?
(a) \( (a+1)x^2 - \frac{3}{5}x = 11 \), where \( a \neq - 1 \)
(b) \( (3 - x)^2 - 5 = x^2 + 2x + 1 \)
(c) \( 8x^3 - x^2 = (2x - 1)^3 \)
(d) \( - 3x^2 = (2-x)(3x-\frac{1}{2}) \)
Answer: (a) (a+1)x^2 - \frac{3}{5}x = 11, where a \neq 1
Explanation : We will check for all the options. For (a), \( (a+1)x^2 - \frac{3}{5}x = 11 \). So, if \( a \neq - 1 \), then coefficient of \( x^2 \) will not be zero. So, it will be a quadratic equation and in rest of 3 options, on expanding coefficient of \( x^2 \) becomes zero.

Question. Which of the following is a solution of quadratic equation \( x^2 - b^2 = a(2x - a) \)?
(a) \( a + b \)
(b) \( 2b - a \)
(c) \( ab \)
(d) \( \frac{a}{b} \)
Answer: (a) a + b

Question. The roots of the quadratic equation \( x^2 - 3x - m(m + 3) = 0 \), where \( m \) is a constant are :
(a) \( m, m + 3 \)
(b) \( - m, m + 3 \)
(c) \( m, - (m + 3) \)
(d) \( - m, - (m + 3) \)
Answer: (b) - m, m + 3

Question. The quadratic equation \( 2y^2 - \sqrt{3}y + 1 = 0 \) has
(a) more than two real roots
(b) two equal real roots
(c) no real roots
(d) two distinct real roots
Answer: (c) no real roots
Explanation : Discriminant of quadratic equation : \( D = b^2 - 4ac \). \( D = 3 - 4(2)(1) = - 5 \). Which is negative. So this quadratic equation has no real roots.

Question. Which one of the following equations has no real roots?
(a) \( x^2 - 2x - 2\sqrt{3} = 0 \)
(b) \( x^2 - 4x + 4\sqrt{2} = 0 \)
(c) \( 3x^2 + 4\sqrt{3}x + 3 = 0 \)
(d) \( x^2 + 4x - 2\sqrt{2} = 0 \)
Answer: (a) x^2 - 2x - 2\sqrt{3} = 0
Explanation : For distinct real roots \( D > 0 \). Discriminant of quadratic equation \( x^2 + 2x - 7 = 0 \). \( b^2 - 4ac = (2)^2 - 4(1)(- 7) = 4 + 28 = 32 \) which is positive. So this equation has distinct roots.

Question. If the equation \( x^2 + 4x + k = 0 \) has real and distinct roots, then
(a) \( k \leq 4 \)
(b) \( k < 4 \)
(c) \( k > 4 \)
(d) \( k \geq 4 \)
Answer: (b) k < 4
Explanation : For real and distinct roots \( D > 0 \). \( b^2 - 4ac > 0 \Rightarrow 16 - 4k > 0 \Rightarrow 16 > 4k \Rightarrow 4k < 16 \Rightarrow k < 4 \)

Question. The quadratic equation \( 49x^2 + 21x + \frac{9}{4} = 0 \) has
(a) real and equal roots
(b) four real roots
(c) real and unequal roots
(d) no real roots
Answer: (a) real and equal roots
Explanation : Discriminant of the equation \( 49x^2 + 21x + \frac{9}{4} = 0 \). \( D = b^2 - 4ac \). \( D = (21)^2 - 4(49)(\frac{9}{4}) = 441 - 441 = 0 \). Here \( D = 0 \). So roots are real and equal.

Question. The general form of a quadratic equation is :
(a) \( ax^2 + bx + c \)
(b) \( ax^2 + bx + c = 0 \)
(c) \( a^2x + b \)
(d) \( ax^2 + bx + c = 0, a \neq 0 \)
Answer: (d) ax^2 + bx + c = 0, a \neq 0
Explanation : By the definition of quadratic equation it is the form \( ax^2 + bx + c, a \neq 0 \).

Question. The number of possible solutions of a quadratic equation are :
(a) exactly two
(b) at most two
(c) at least two
(d) None of the options
Answer: (b) at most two
Explanation : A quadratic equation cannot have more than 2 solution i.e., either 2, 1 or 0.

Question. The roots of \( 3x^2 - 7x + 4 = 0 \) are :
(a) rationals
(b) irrationals
(c) Positive integers
(d) negative integers
Answer: (a) rationals
 Explanation : Discriminant of the equation \( 3x^2 - 7x + 4 = 0 \). \( D = 49 - 48 = 1 \) which is a perfect square. So roots are rationals.

Question. The roots of equation \( x + \frac{16}{x} = 10 \) are :
(a) 4, 6
(b) 4, 4
(c) 4, 5
(d) 2, 8
Answer: (d) 2, 8

Question. If a, b are the roots of \( x^2 + px + q = 0 \), then the value of \( \frac{a}{b} + \frac{b}{a} \) is :
(a) \( \frac{p^2 - 2q}{q} \)
(b) \( \frac{2q - p^2}{q} \)
(c) \( \frac{p^2 + 2q}{q} \)
(d) None of the options
Answer: (a) \frac{p^2 - 2q}{q}
Explanation : Here, \( a + b = - p \) and \( ab = q \). \( \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} = \frac{(a + b)^2 - 2ab}{ab} = \frac{p^2 - 2q}{q} \)

Question. If the roots of \( ax^2 + bx + c = 0 \) be equal, then the value of \( c \) is :
(a) \( -\frac{b}{2a} \)
(b) \( \frac{b}{2a} \)
(c) \( -\frac{b^2}{4a} \)
(d) \( \frac{b^2}{4a} \)
Answer: (d) \frac{b^2}{4a}
Explanation : Roots are equal so \( b^2 - 4ac = 0 \Rightarrow b^2 = 4ac \Rightarrow c = \frac{b^2}{4a} \)

Question. If the sum of the roots of an equation is 6 and one root is \( 3 - \sqrt{5} \), then the equation is :
(a) \( x^2 - 6x + 4 = 0 \)
(b) \( x^2 - 4x + 6 = 0 \)
(c) \( x^2 - 6x + 5 = 0 \)
(d) None of the options
Answer: (a) x^2 - 6x + 4 = 0
Explanation : Here \( a + b = 6 \) and \( a = 3 - \sqrt{5} \). So, \( b = 3 + \sqrt{5} \). Now, \( ab = (3 - \sqrt{5})(3 + \sqrt{5}) = 4 \). Required quadratic equation is \( x^2 - 6x + 4 = 0 \)

Question. If a, b be the roots of \( ax^2 + bx + c = 0 \), then the value of \( a^2 + b^2 \) is :
(a) \( \frac{b^2 - 2ac}{2a} \)
(b) \( \frac{b^2 - 4ac}{2a} \)
(c) \( \frac{b^2 - 2ac}{a^2} \)
(d) \( \frac{b^2 + 4ac}{2a} \)
Answer: (c) \frac{b^2 - 2ac}{a^2}
Explanation : \( a^2 + b^2 = (a + b)^2 - 2ab = (\frac{-b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2} \)