CBSE Class 10 Mathematics Quadratic Equations MCQs Set 18

Answer: (a) 3

Explanation:
1. Put \( y = 1 \) in the first equation: \( a + a + 3 = 0 \), so \( a = -\frac{3}{2} \).
2. Put \( y = 1 \) in the second equation: \( 1 + 1 + b = 0 \), so \( b = -2 \).
3. \( ab = \left(-\frac{3}{2}\right)(-2) = 3 \). The answer is option (a).

Teacher's Note:
1. A common root must satisfy both equations, so substitute it in each one.
2. Solve for a and b separately and then multiply.

Answer: (a) real and unequal

Explanation:
1. \( D = b^2 - 4ac = (-3)^2 - 4(1)(-9) = 9 + 36 = 45 \).
2. Since \( D > 0 \), the roots are real and unequal.
3. Also 45 is not a perfect square, so the roots are irrational (not rational). The answer is option (a).

Teacher's Note:
1. Find the discriminant first; its sign decides the nature of the roots.
2. Rational roots need D to be a perfect square.

Answer: (a) real and unequal

Explanation:
1. Multiply by \( x \) (x is not 0): \( x^2 - 18 = 6x \), i.e. \( x^2 - 6x - 18 = 0 \).
2. \( D = (-6)^2 - 4(1)(-18) = 36 + 72 = 108 > 0 \), so the roots are real and unequal.
3. 108 is not a perfect square, so the roots are irrational, not rational. The answer is option (a).

Teacher's Note:
1. Clear the fraction first to get a quadratic equation in standard form.
2. A positive discriminant that is not a perfect square gives irrational roots.

Answer: (a) \( \pm 0.2 \)

Explanation:
1. \( x^2 = 0.04 = (0.2)^2 \).
2. \( (x + 0.2)(x - 0.2) = 0 \), so \( x = \pm 0.2 \). The answer is option (a).

Teacher's Note:
1. This is a difference of two squares: a² - b² = (a + b)(a - b).
2. Do not forget the negative root.

Answer: (c) -5

Explanation:
1. For \( ax^2 + bx + c \) the sum of the zeroes is \( -\frac{b}{a} \).
2. Here \( \alpha + \beta = -\frac{5}{1} = -5 \). The answer is option (c).

Teacher's Note:
1. Sum of zeroes = -(coefficient of x) / (coefficient of x²).
2. The sum is -5 even though the zeroes themselves are not real.

Answer: (a) 40

Explanation:
1. \( D = b^2 - 4ac = (-4)^2 - 4(3)(-2) = 16 + 24 = 40 \). The answer is option (a).

Teacher's Note:
1. Take care with signs: -4ac becomes +24 here.
2. Write a, b, c clearly before substituting.

Answer: (b) -2

Explanation:
1. Since 0.5 is a root, it satisfies the equation: \( 2(0.5)^2 + 3(0.5) + c = 0 \).
2. \( 0.5 + 1.5 + c = 0 \), so \( c = -2 \). The answer is option (b).

Teacher's Note:
1. A root of an equation always satisfies it, so substitute directly.
2. Do the arithmetic in decimals carefully.

Answer: (d) \( x^2 - 4x + 16 = 0 \)

Explanation:
1. For \( x^2 - 2x + 4 = 0 \), the sum of roots \( \alpha + \beta = 2 \) and the product \( \alpha\beta = 4 \).
2. New roots \( 2\alpha, 2\beta \) have sum \( 4 \) and product \( 4 \times 4 = 16 \).
3. The equation is \( x^2 - 4x + 16 = 0 \), option (d).

Teacher's Note:
1. Use x² - (sum)x + (product) = 0 to build the new equation.
2. Doubling each root doubles the sum and multiplies the product by 4.

Answer: (a) 4.5

Explanation:
1. Factorise: \( 2x^2 - 10x - x + 5 = 0 \) gives \( (2x - 1)(x - 5) = 0 \).
2. The roots are \( \frac{1}{2} \) and 5, and their difference is \( 5 - \frac{1}{2} = 4.5 \). The answer is option (a).

Teacher's Note:
1. Find the roots by splitting the middle term.
2. Alternatively use (alpha - beta)² = (alpha + beta)² - 4 alpha beta.

Answer: (b) 7

Explanation:
1. Real roots need \( D \ge 0 \), i.e. \( a^2 \ge 4b \).
2. For \( b = 1 \): \( a^2 \ge 4 \), so a = 2, 3, 4 (3 pairs).
3. For \( b = 2 \): \( a^2 \ge 8 \), so a = 3, 4 (2 pairs).
4. For \( b = 3 \): \( a^2 \ge 12 \), so a = 4 (1 pair). For \( b = 4 \): \( a^2 \ge 16 \), so a = 4 (1 pair).
5. Total pairs = 3 + 2 + 1 + 1 = 7: option (b).

Teacher's Note:
1. Real roots means D is greater than or equal to 0, including the equal-roots case.
2. Go through each value of b systematically so that no pair is missed.

Answer: (c) \( \sqrt{2} \)

Explanation:
1. From the first equation \( y^2 = x^2 \). Put this in the second: \( (x - k)^2 + x^2 = 1 \), i.e. \( 2x^2 - 2kx + k^2 - 1 = 0 \).
2. A unique value of x needs \( D = 0 \): \( 4k^2 - 8(k^2 - 1) = 0 \), so \( k^2 = 2 \) and \( k = \pm\sqrt{2} \).
3. Then \( x = \frac{k}{2} \), which is positive only for \( k = \sqrt{2} \): option (c).

Teacher's Note:
1. Eliminate y by using y² = x².
2. A unique solution of a quadratic in x means a zero discriminant; then check the sign condition.

Answer: (d) A and C are of opposite sign

Explanation:
1. Sum of roots \( = \frac{B}{A} = -1 + 1 = 0 \), so \( B = 0 \).
2. Product of roots \( = \frac{C}{A} = (-1)(1) = -1 \), so \( C = -A \).
3. Hence A and C have opposite signs: option (d).

Teacher's Note:
1. Use the sum and product of roots for Ax² - Bx + C = 0: sum = B/A and product = C/A.
2. A quadratic needs A to be non-zero.

Answer: (d) \( a^3 + b^3 + c^3 = 3abc \)

Explanation:
1. Equal roots mean \( D = 0 \): \( 4(b^2 - ac)^2 - 4(a^2 - bc)(c^2 - ab) = 0 \).
2. Expanding, \( b^4 - 2ab^2c + a^2c^2 - a^2c^2 + a^3b + bc^3 - ab^2c = 0 \), which simplifies to \( b(a^3 + b^3 + c^3 - 3abc) = 0 \).
3. Since \( b \ne 0 \), \( a^3 + b^3 + c^3 = 3abc \): option (d).

Teacher's Note:
1. Expand carefully; many terms cancel.
2. The condition b is not 0 lets you divide by b at the end.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. Let the roots be \( \alpha \) and \( 2\alpha \). Sum \( 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \) and product \( 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \).
2. Eliminating \( \alpha \) gives \( 2(\text{sum})^2 = 9(\text{product}) \), so \( (3a - 1)^2 = 9(a^2 - 5a + 3) \).
3. \( 9a^2 - 6a + 1 = 9a^2 - 45a + 27 \) gives \( 39a = 26 \), so \( a = \frac{2}{3} \). Check: the equation becomes \( x^2 + 9x + 18 = 0 \) with roots -3 and -6. Option (a).

Teacher's Note:
1. For roots alpha and 2 alpha, the condition 2(sum)² = 9(product) saves a lot of work.
2. Always verify by substituting the value back.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. Put \( x = 3 \): \( 9k - 3k - 3 = 0 \), so \( 6k = 3 \) and \( k = \frac{1}{2} \). The answer is option (b).

Teacher's Note:
1. Substitute the given root and solve the resulting linear equation in k.
2. Check: (1/2)(9) - (1/2)(3) - 3 = 0.

Answer: (b) 0, -7

Explanation:
1. Put \( x = 2 \): \( 4 + 2(a + 1) + b + 1 = 0 \), so \( 2a + b = -7 \) (i).
2. Put \( x = -3 \): \( 9 - 3(a + 1) + b + 1 = 0 \), so \( -3a + b = -7 \) (ii).
3. Subtract (ii) from (i): \( 5a = 0 \), so \( a = 0 \) and \( b = -7 \). The answer is option (b).

Teacher's Note:
1. Each root gives one linear equation in a and b.
2. Check with sum and product: sum = -1 = -(a + 1) and product = -6 = b + 1.

Answer: (d) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^2 \)

Explanation:
1. Expand option (d): \( x^4 + 4x^3 + 4x^2 = x^4 + 3 + 4x^2 \), which gives \( 4x^3 - 3 = 0 \), a cubic equation, so it is not quadratic.
2. Option (a) simplifies to \( 2x^2 - 6x - 1 = 0 \) and (b) to \( 2x^2 - 2x + 5 = 0 \), both quadratic. In option (c), \( (\sqrt{2} + \sqrt{3})^2 x^2 + x^2 - 3x^2 + 5x = 0 \) gives \( (3 + 2\sqrt{6})x^2 + 5x = 0 \), also quadratic.

Teacher's Note:
1. Simplify each equation to the form p(x) = 0 and look at the highest power of x.
2. If the x² coefficient cancels out or a higher power remains, it is not quadratic.

Answer: (c) \( 2x^2 - 7x + 6 = 0 \)

Explanation:
1. Substitute \( x = 2 \) in each equation.
2. (a) \( 4 - 8 + 5 = 1 \); (b) \( 4 + 6 - 12 = -2 \); (c) \( 8 - 14 + 6 = 0 \); (d) \( 12 - 12 - 2 = -2 \).
3. Only option (c) gives zero.

Teacher's Note:
1. A number is a root only if it makes the expression equal to zero.
2. Check every option; do not stop at the first non-zero value.

Answer: (a) \( (a + 1)x^2 - \frac{3}{5}x = 11 \), where \( a \ne -1 \)

Explanation:
1. In option (a) the coefficient of \( x^2 \) is \( a + 1 \ne 0 \), so it is quadratic.
2. In (b), expanding gives \( x^2 - 6x + 4 = x^2 + 2x + 1 \), so \( -8x + 3 = 0 \) is linear.
3. In (c), \( 8x^3 - x^2 = x^3 - 3x^2 + 3x - 1 \) is a cubic. In (d), the right side expands to \( -3x^2 + \frac{13}{2}x - 1 \), so the \( x^2 \) terms cancel and the equation is linear.
4. Only option (a) is quadratic.

Teacher's Note:
1. The x² coefficient must not vanish; check this after expanding.
2. Cubic and linear cases appear when higher or lower powers survive.

Answer: (a) \( a + b \)

Explanation:
1. Rewrite: \( x^2 - 2ax + a^2 - b^2 = 0 \), i.e. \( (x - a)^2 - b^2 = 0 \).
2. \( (x - a - b)(x - a + b) = 0 \), so \( x = a + b \) or \( x = a - b \).
3. Only \( a + b \) appears among the options: option (a).

Teacher's Note:
1. Collect everything on the left and spot the perfect square (x - a)².
2. Then use the difference of squares.

Answer: (b) \( -m, m + 3 \)

Explanation:
1. Split the middle term: \( x^2 - (m + 3)x + mx - m(m + 3) = 0 \).
2. \( x\{x - (m + 3)\} + m\{x - (m + 3)\} = 0 \), so \( (x + m)(x - (m + 3)) = 0 \).
3. Hence \( x = -m \) or \( x = m + 3 \): option (b).

Teacher's Note:
1. Write 3 as (m + 3) - m to split the middle term.
2. Check: the sum of roots is (-m) + (m + 3) = 3.

Answer: (d) \( ax^2 + bx + c = 0, a \ne 0 \)

Explanation:
1. A quadratic equation is an equation of degree 2 written as \( ax^2 + bx + c = 0 \) with real numbers a, b, c.
2. The condition \( a \ne 0 \) is essential; otherwise the equation becomes linear. Hence option (d).

Teacher's Note:
1. An expression without '= 0' is a polynomial, not an equation.
2. Always mention a is not equal to 0.

Answer: (b) at most two

Explanation:
1. A quadratic equation can have two distinct real roots, one repeated real root, or no real root.
2. It can never have more than two roots, so the number of solutions is at most two: option (b).

Teacher's Note:
1. The degree of a polynomial equation limits the number of its roots.
2. Zero real roots occurs when D is negative.

Answer: (a) \( \frac{p^2 - 2q}{q} \)

Explanation:
1. \( \alpha + \beta = -p \) and \( \alpha\beta = q \).
2. \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{p^2 - 2q}{q} \). Option (a).

Teacher's Note:
1. Convert the expression into one that uses only the sum and product of roots.
2. Remember alpha² + beta² = (alpha + beta)² - 2 alpha beta.

Answer: (a) \( x^2 - 6x + 4 = 0 \)

Explanation:
1. The other root is \( 6 - (3 - \sqrt{5}) = 3 + \sqrt{5} \).
2. Product of roots \( = (3 - \sqrt{5})(3 + \sqrt{5}) = 9 - 5 = 4 \).
3. The equation is \( x^2 - 6x + 4 = 0 \): option (a).

Teacher's Note:
1. Find the second root from the given sum.
2. Use x² - (sum)x + (product) = 0.

Answer: (a) \( ax^2 - (a^2 + 1)x + a = 0 \)

Explanation:
1. Sum of roots \( = a + \frac{1}{a} = \frac{a^2 + 1}{a} \) and product \( = a \times \frac{1}{a} = 1 \).
2. \( x^2 - \frac{a^2 + 1}{a}x + 1 = 0 \). Multiplying by \( a \) gives \( ax^2 - (a^2 + 1)x + a = 0 \): option (a).

Teacher's Note:
1. Multiply through by a to clear the fraction.
2. Check: put x = a, a(a²) - (a² + 1)a + a = 0.

Answer: (a) k = 4 only

Explanation:
1. Write it as \( kx^2 - 3kx + 9 = 0 \). Then \( D = 9k^2 - 36k = 9k(k - 4) \).
2. D = 0 gives \( k = 0 \) or \( k = 4 \). But for \( k = 0 \) the equation reduces to \( 9 = 0 \), which is not a quadratic equation.
3. So the only valid value is \( k = 4 \): option (a).

Teacher's Note:
1. The coefficient of x² must be non-zero for the equation to be quadratic.
2. Reject any value of k that makes the equation lose its x² term.

Answer: (c) both (a) and (b)

Explanation:
1. Let the number be \( x \): \( x + \frac{1}{x} = \frac{10}{3} \), so \( 3x^2 - 10x + 3 = 0 \).
2. \( 3x^2 - 9x - x + 3 = 0 \) gives \( (3x - 1)(x - 3) = 0 \), so \( x = 3 \) or \( x = \frac{1}{3} \).
3. Both values satisfy the condition: option (c).

Teacher's Note:
1. A number and its reciprocal give the same pair of values.
2. Check: 3 + 1/3 = 10/3.

Answer: (b) \( b - c = 0 \)

Explanation:
1. For \( ax^2 - bx + c = 0 \) the sum of roots is \( \frac{b}{a} \) and the product is \( \frac{c}{a} \).
2. Equal sum and product: \( \frac{b}{a} = \frac{c}{a} \), so \( b = c \), i.e. \( b - c = 0 \): option (b).

Teacher's Note:
1. Note the sign: the equation has -bx, so the sum of roots is +b/a.
2. Equate the two expressions and cancel a.

Answer: (c) 5, -3

Explanation:
1. Equal roots mean \( D = 0 \): \( (a + 1)^2 - 4b^2 = 0 \), so \( a + 1 = \pm 2b \).
2. Test the options: (a) \( 36 \ne 16 \); (b) \( 16 \ne 64 \); (c) \( a = 5, b = -3: 36 = 36 \); (d) \( 36 \ne 64 \).
3. Only option (c) satisfies the condition.

Teacher's Note:
1. Use D = 0 and test each pair from the options.
2. b can be negative because only b² appears.

Answer: (a) \( x = 0, -1 \)

Explanation:
1. Subtract 1 from both sides: \( x^2 + x = 0 \).
2. \( x(x + 1) = 0 \), so \( x = 0 \) or \( x = -1 \): option (a).

Teacher's Note:
1. Simplify first to reach standard form.
2. Never divide by x; factorise to keep the root x = 0.

Answer: (c) \( \frac{-5 + \sqrt{21}}{2}, \frac{-5 - \sqrt{21}}{2} \)

Explanation:
1. Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 4}}{2} = \frac{-5 \pm \sqrt{21}}{2} \). Option (c).

Teacher's Note:
1. Compute D = 25 - 4 = 21 first.
2. Check the sum of roots: it must be -5.

Answer: (c) 13

Explanation:
1. Let the son's age be \( x \) years; the father's age is \( (x + 25) \) years.
2. \( x(x + 25) = 84 \) gives \( x^2 + 25x - 84 = 0 \), i.e. \( (x + 28)(x - 3) = 0 \).
3. Age cannot be negative, so \( x = 3 \). After 10 years the son will be \( 3 + 10 = 13 \) years old: option (c).

Teacher's Note:
1. Reject the negative root because an age cannot be negative.
2. Read the question again: it asks for the age after 10 years, not now.

Answer: (d) real and unequal

Explanation:
1. \( D = (\sqrt{5})^2 - 4(2)(-1) = 5 + 8 = 13 > 0 \).
2. A positive discriminant means real and unequal roots: option (d).

Teacher's Note:
1. Square the coefficient carefully: (√5)² = 5.
2. When a and c have opposite signs, D is always positive.

Answer: (b) \( k < 4 \)

Explanation:
1. \( D = 16 - 4k > 0 \) for real and distinct roots.
2. So \( 4k < 16 \), i.e. \( k < 4 \): option (b).

Teacher's Note:
1. Distinct roots need D > 0 strictly (not equal to 0).
2. Dividing by a positive number keeps the inequality direction.

Answer: (a) real and equal roots

Explanation:
1. \( D = 21^2 - 4(49)\left(\frac{9}{4}\right) = 441 - 441 = 0 \).
2. Since \( D = 0 \), the roots are real and equal: option (a).

Teacher's Note:
1. Compute 4ac as 4 x 49 x 9/4 = 441.
2. A quadratic equation can never have four roots.

Answer: (d) 8

Explanation:
1. Let the roots be \( \alpha \) and \( 2\alpha \). Sum: \( 3\alpha = \frac{18}{9} = 2 \), so \( \alpha = \frac{2}{3} \).
2. Product: \( 2\alpha^2 = \frac{k}{9} \), so \( 2 \times \frac{4}{9} = \frac{k}{9} \) and \( k = 8 \): option (d).

Teacher's Note:
1. Use roots alpha and 2 alpha with the sum and product formulas.
2. Check: 9x² - 18x + 8 = (3x - 2)(3x - 4) has roots 2/3 and 4/3.

Answer: (d) All of the above

Explanation:
1. If \( a\alpha^2 + b\alpha + c = 0 \), then \( \alpha \) itself (not \( -\alpha \)) is a solution, so statement (a) is incorrect.
2. The roots of \( ax^2 + bx + c = 0 \) are exactly the zeroes of the polynomial, not their additive inverses, so statement (b) is incorrect.
3. The other root is \( \frac{c}{a\alpha} \) (or \( -\frac{b}{a} - \alpha \)), not always \( -\alpha \), so statement (c) is incorrect.
4. All three statements are incorrect, so the answer is option (d).

Teacher's Note:
1. A root is a value of x that makes the expression zero; it is the same as a zero of the polynomial.
2. The other root depends on the equation, for example x² - 5x + 6 = 0 has roots 2 and 3.

Answer: (c) A is true, but R is false.

Explanation:
1. The roots are \( \frac{-a \pm \sqrt{a^2 - 4b}}{2} \). If they are rational, \( a^2 - 4b \) is a perfect square, say \( m^2 \), and \( m \) is an integer.
2. Since \( a^2 - 4b = m^2 \) and a, m have the same parity, \( x = \frac{-a \pm m}{2} \) is an integer. So the Assertion is true.
3. The Reason is false: for example \( x^2 + 2x - 1 = 0 \) has unit coefficient but irrational roots \( -1 \pm \sqrt{2} \). Hence option (c).

Teacher's Note:
1. A monic equation with integer coefficients and a rational root has an integer root (rational root theorem).
2. Test a Reason with a quick counter-example.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( f(1) + f(2) = 0 \) gives \( 5a + 3b + 2c = 0 \) (1). Since -1 is a root, \( a - b + c = 0 \) (2).
2. From (2), \( c = b - a \). Put in (1): \( 5a + 3b + 2b - 2a = 0 \), so \( 3a + 5b = 0 \) and \( b = -\frac{3a}{5} \), \( c = -\frac{8a}{5} \).
3. Product of roots \( \frac{c}{a} = -\frac{8}{5} \), so the other root is \( \frac{-8/5}{-1} = \frac{8}{5} \). The Assertion is true.
4. The Reason is the standard sum and product of roots, which is true and was used in the calculation, but it is a general formula and does not by itself give the value 8/5. Hence option (b).

Teacher's Note:
1. Form two equations in a, b, c from the given conditions.
2. Compare how much the Reason directly explains the Assertion.