If x = 3 is a root of x2 - kx + 6 = 0, find k and the other root.

9 - 3k + 6 = 0, so k = 5. The equation is x2 - 5x + 6 = (x - 2)(x - 3) = 0, so the other root is 2.

Solve by factorisation: 3x2 - 11x - 4 = 0.

ac = -12; the numbers are -12 and 1. 3x2 - 12x + x - 4 = 3x(x - 4) + 1(x - 4) = (x - 4)(3x + 1) = 0, so x = 4 or x = -(1)/(3).

A student solves x2 = 6x by dividing both sides by x and gets x = 6. What has gone wrong?

Dividing by x assumes x ≠ 0 and loses a root. Correctly, x2 - 6x = x(x - 6) = 0, so x = 0 or x = 6.

Quadratic Equations

Q: Define a quadratic equation in one variable. Why must the coefficient of x2 be non-zero?

A quadratic equation in one variable is an equation that can be written as ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0. If a = 0, the x2 term disappears and the equation becomes linear.

Q: Is (x - 3)(2x + 1) = x(x + 5) a quadratic equation? Give a reason.

Expanding gives 2x2 - 5x - 3 = x2 + 5x, so x2 - 10x - 3 = 0. The x2 term survives with coefficient 1, so the equation is quadratic.

Q: Solve by factorisation: x2 + 5x - 24 = 0.

(x + 8)(x - 3) = 0, so x = -8 or x = 3.

Q: Solve 2x2 - 7x + 6 = 0 for x in N.

(2x - 3)(x - 2) = 0 gives x = (3)/(2) or x = 2. Only 2 is a natural number, so the solution set is \2\.

Q: Solve x2 - 6x + 4 = 0, giving the roots correct to two decimal places.

D = 36 - 16 = 20, so x = (6 ± 2√(5))/(2) = 3 ± √(5). With √(5) = 2.2361, x = 5.24 or x = 0.76.

ICSE Class 10 · Mathematics · Chapter 5

What this chapter covers

What a quadratic equation is
Roots of a quadratic equation
Solving quadratic equations by factorisation
Solving quadratic equations using the formula
The discriminant
Nature of roots
Finding an unknown from the nature of roots
Choosing a method
Common errors and how to avoid them

Key points

A quadratic equation in one variable has the standard form ax² + bx + c = 0 with a ≠ 0; expand and simplify before deciding whether an equation is quadratic.
A root is a value of x that satisfies the equation; a quadratic equation has at most two roots.
Factorisation uses the rule that if pq = 0 then p = 0 or q = 0, so one side must be 0 first.
To split the middle term of ax² + bx + c, find two numbers with product ac and sum b.
Never divide by x; take it out as a common factor so the root x = 0 is kept.
The quadratic formula is x = (-b ± √(b² - 4ac))/(2a); work out b² - 4ac first and round only at the end.
The discriminant is D = b² - 4ac.
D > 0: two distinct real roots; D = 0: two equal real roots, each -(b)/(2a); D < 0: no real roots.
For an unknown constant, use D = 0 for equal roots, D > 0 for distinct real roots, D ≥ 0 for real roots and D < 0 for no real roots.
Reject any root that makes a denominator zero or lies outside the stated replacement set.

Questions answered

Define a quadratic equation in one variable. Why must the coefficient of x² be non-zero?

A quadratic equation in one variable is an equation that can be written as ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. If a = 0, the x² term disappears and the equation becomes linear.

Is (x - 3)(2x + 1) = x(x + 5) a quadratic equation? Give a reason.

Expanding gives 2x² - 5x - 3 = x² + 5x, so x² - 10x - 3 = 0. The x² term survives with coefficient 1, so the equation is quadratic.

If x = 3 is a root of x² - kx + 6 = 0, find k and the other root.

9 - 3k + 6 = 0, so k = 5. The equation is x² - 5x + 6 = (x - 2)(x - 3) = 0, so the other root is 2.

Solve by factorisation: x² + 5x - 24 = 0.

(x + 8)(x - 3) = 0, so x = -8 or x = 3.

Solve by factorisation: 3x² - 11x - 4 = 0.

ac = -12; the numbers are -12 and 1. 3x² - 12x + x - 4 = 3x(x - 4) + 1(x - 4) = (x - 4)(3x + 1) = 0, so x = 4 or x = -(1)/(3).

A student solves x² = 6x by dividing both sides by x and gets x = 6. What has gone wrong?

Dividing by x assumes x ≠ 0 and loses a root. Correctly, x² - 6x = x(x - 6) = 0, so x = 0 or x = 6.

Solve 2x² - 7x + 6 = 0 for x in N.

(2x - 3)(x - 2) = 0 gives x = (3)/(2) or x = 2. Only 2 is a natural number, so the solution set is \2\.

Solve x² - 6x + 4 = 0, giving the roots correct to two decimal places.

D = 36 - 16 = 20, so x = (6 ± 2√(5))/(2) = 3 ± √(5). With √(5) = 2.2361, x = 5.24 or x = 0.76.

Past board exam questions

ICSE 2027 sample paper · 1 mark

If the roots of the quadratic equation x² - 8x + p² = 0 are equal, then the positive value of p is:

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ICSE 2027 sample paper · 4 marks

Solve the quadratic equation (x - 2)² + 11x - 7 = 0 and give your answer correct to 3 significant figures.

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(x - 2)² + 11x - 7 = 0 ⇒ x² - 4x + 4 + 11x - 7 = 0 ⇒ x² + 7x - 3 = 0 x = (-7 ± √(49 - 4(1)(-3)))/(2) = (-7 ± √(61))/(2) = (-7 ± 7.8102)/(2) x = 0.4051 or x = -7.4051 x = 0.405 or x = -7.41 (to 3 significant figures)

ICSE 2027 sample paper · 3 marks

A rectangle measures 12 m by 10 m. The shaded region is a strip of uniform width x m along its top side and its right side (an L-shape), leaving an unshaded rectangle in the bottom-left corner. The area of the shaded region is 57 m². Form an equation in x and solve it to find the possible value of x.

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Unshaded rectangle = (12 - x)(10 - x). 12 × 10 - (12 - x)(10 - x) = 57 120 - (120 - 22x + x²) = 57 ⇒ x² - 22x + 57 = 0 (x - 19)(x - 3) = 0 ⇒ x = 19 or x = 3 x = 19 is not possible (it is more than 10 m), so x = 3 m.

ICSE 2026 board paper · 1 mark

The nature of roots of quadratic equation 3x² - 6x - 3 = 0 are:

real and equal
real, distinct and rational
real, distinct and irrational
no real roots

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real, distinct and irrational

ICSE 2026 sample paper · 1 mark

The roots of quadratic equation x² - 1 = 0 are:

0, 0
1, 1
-1, -1
+1, -1

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+1, -1

ICSE 2026 sample paper · 4 marks

The roots of equation (q - r)x² + (r - p)x + (p - q) = 0 are equal. Prove that: 2q = p + r, that is, p, q & r are in A.P.

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Here a = q − r, b = r − p, c = p − q. Equal roots ⟹ b² = 4ac: (r - p)² = 4(q - r)(p - q) r² + p² - 2pr = 4(pq - q² - pr + qr) r² + p² + 2pr = 4pq - 4q² + 4qr (p + r)² - 4q(p + r) + 4q² = 0 (p + r - 2q)² = 0 ⇒ p + r = 2q. Hence 2q = p + r, i.e. p, q, r are in A.P.

ICSE 2026 sample paper · 3 marks

The sum of the squares of three consecutive even numbers is 596. Find the numbers.

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Let the numbers be x − 2, x, x + 2. (x - 2)² + x² + (x + 2)² = 596 ⇒ 3x² + 8 = 596 ⇒ x² = 196 ⇒ x = ± 14. The numbers are 12, 14 and 16 (taking natural numbers; x = −14 gives −16, −14, −12).

ICSE 2026 board paper · 4 marks

Solve the quadratic equation (x - 2)² - 5x - 3 = 0 and give your answer correct to 3 significant figures. (Use Mathematical Tables for this question if necessary.)

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x² - 4x + 4 - 5x - 3 = 0 ⇒ x² - 9x + 1 = 0, so a = 1, b = -9, c = 1. D = 81 - 4 = 77, √(77) = 8.775 x = (9 ± 8.775)/(2) x = (17.775)/(2) = 8.8875 ≈ 8.89 or x = (0.225)/(2) = 0.1125 ≈ 0.113 x = 8.89 or 0.113 (correct to 3 significant figures).