Quadratic Equations — Chapter Test | Class 10 IMO

Q1

If a quadratic has roots that are reciprocals of each other, then:

Product of roots = c/a = 1 means a = c.

Q2

The discriminant of 2x² − 4x + 1 = 0 is:

D = b² − 4ac = 16 − 8 = 8.

Q3

The roots of x² − 7x + 12 = 0 are:

x² − 7x + 12 = (x − 3)(x − 4), so the roots are 3 and 4.

Q4

If the discriminant is negative, the equation has:

D < 0 means the roots are not real.

Q5

A positive number added to its reciprocal gives 10/3. The number is:

x + 1/x = 10/3 gives 3x² − 10x + 3 = 0 = (3x − 1)(x − 3), so x = 3 (or 1/3).

Q6

For x² − kx + 4 = 0 to have equal roots, k equals:

Equal roots need k² − 16 = 0, so k = ±4.

Q7

The hypotenuse of a right triangle is 13 cm and one side is 7 cm more than the other. The shorter side is:

x² + (x + 7)² = 169 gives 2x² + 14x − 120 = 0, i.e. x² + 7x − 60 = (x + 12)(x − 5) = 0, so x = 5 cm.

Q8

If the sum of the roots of a quadratic is 0, the equation has the form:

Sum = −b/a = 0 forces b = 0, leaving ax² + c = 0.

Q9

The sum of the squares of the roots of x² − 5x + 6 = 0 is:

Sum = 5, product = 6, so sum of squares = 25 − 12 = 13.

Q10

The value of √(6 + √(6 + √(6 + …))) is:

Let x = √(6 + x); then x² = 6 + x gives x² − x − 6 = (x − 3)(x + 2) = 0, so x = 3.

Q11

Which equation has roots of opposite signs?

x² − x − 6 = 0 has product −6 < 0, so the roots (3 and −2) have opposite signs.

Q12

For what value of k (k ≠ 12) does (k − 12)x² + 2(k − 12)x + 2 = 0 have equal roots?

D = 0: 4(k − 12)² − 8(k − 12) = 0 gives 4(k − 12)(k − 14) = 0, so k = 14.

Q13

If both roots of a quadratic are equal to 4, the equation could be:

(x − 4)² = x² − 8x + 16 = 0 has the double root 4.

Q14

If one root of x² − 5x + k = 0 is 2, then k equals:

Substituting x = 2: 4 − 10 + k = 0, so k = 6.

Q15

The area of a right triangle is 24 cm² and its base is 2 cm more than its height. The height is:

½·h·(h + 2) = 24 gives h² + 2h − 48 = (h + 8)(h − 6) = 0, so h = 6 cm (base 8 cm).