Quadratic equations: Class 10 Mathematics Notes, Short Questions and Numericals

The standard form, solving by factorisation, by completing the square and by the quadratic formula, equations that reduce to quadratics, and word problems.

Key points

A quadratic equation in x is an equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The highest power of x is 2, so the equation has at most two roots (solutions). A root is a value of x that makes the equation true.

Method 1: factorisation. Use it when the quadratic expression can be factorised.

Write the equation in the standard form with zero on the right.
Factorise the left side into two brackets.
Use the rule that if A × B = 0 then A = 0 or B = 0, and solve each linear equation.
Check each root by substituting it in the original equation.

Method 2: completing the square.

Divide every term by a so that the coefficient of x² is 1, and move the constant to the right side.
Add the square of half the coefficient of x to both sides.
Write the left side as a perfect square, (x + p)² = q.
Take the square root of both sides, remembering both the positive and the negative root, and solve for x.

Method 3: the quadratic formula. For ax² + bx + c = 0 the roots are x = (−b ± √(b² − 4ac)) / 2a. It is found by completing the square on the general equation and it works for every quadratic equation. Write down a, b and c with their signs before substituting.

The expression b² − 4ac is called the discriminant. If it is positive there are two different real roots, if it is zero there is one repeated root and if it is negative there are no real roots. The next chapter studies it in detail.

Equations reducible to quadratic form. An equation such as x⁴ − 13x² + 36 = 0 can be solved by putting y = x², which gives y² − 13y + 36 = 0. After finding y, return to x using x = ±√y.

Word problems. Choose a letter for the unknown, write the information as an equation, solve it and reject any root that does not fit the problem, for example a negative length. Always state the answer with its unit.

Study tip: Always check the roots by putting them back into the original equation. For a word problem, say which root you reject and why.

Definitions and flashcards

Say the meaning first, then open the card to check.

Quadratic equation

An equation that can be written as ax² + bx + c = 0 with a ≠ 0.

Root

A value of the variable that satisfies the equation. It is also called a solution.

Standard form

The form ax² + bx + c = 0, with all terms on the left side and zero on the right.

Completing the square

Writing a quadratic expression in the form (x + p)² + q by adding and subtracting the right number.

Quadratic formula

The formula x = (−b ± √(b² − 4ac)) / 2a that gives the roots of ax² + bx + c = 0.

Discriminant

The quantity b² − 4ac, which shows the nature of the roots.

Reducible equation

An equation that becomes a quadratic after a substitution such as y = x².

Formula sheet

Name	Formula	What it means
Standard form	ax² + bx + c = 0, a ≠ 0	Every quadratic equation can be written like this.
Quadratic formula	x = (−b ± √(b² − 4ac)) / 2a	Gives both roots.
Completing the square	x² + 2px + p² = (x + p)²	The number added is the square of half of the coefficient of x.
Discriminant	D = b² − 4ac	D > 0: two real roots. D = 0: one repeated root. D < 0: no real roots.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Define a quadratic equation and write its standard form. (2 marks)

A quadratic equation is a polynomial equation in one variable whose highest power is 2. Its standard form is ax² + bx + c = 0, where a ≠ 0.

Why must a be non-zero in ax² + bx + c = 0? (2 marks)

If a = 0 the term in x² disappears and the equation becomes bx + c = 0, which is linear and not quadratic.

State the quadratic formula and name the part under the square root. (2 marks)

The formula is x = (−b ± √(b² − 4ac)) / 2a. The expression b² − 4ac under the square root is the discriminant.

Solve x² − 9x + 20 = 0 by factorisation. (3 marks)

Find two numbers with product 20 and sum −9, which are −4 and −5. So (x − 4)(x − 5) = 0. Then x − 4 = 0 or x − 5 = 0, and x = 4 or x = 5.

Explain why x⁴ − 5x² + 4 = 0 can be solved as a quadratic equation. (3 marks)

Put y = x². Then x⁴ = y² and the equation becomes y² − 5y + 4 = 0, which is quadratic in y. It gives (y − 1)(y − 4) = 0, so y = 1 or y = 4. Since y = x², x = ±1 or x = ±2.

Solve x² + 4x − 5 = 0 by completing the square. (5 marks)

Write x² + 4x = 5. Half of 4 is 2 and its square is 4, so add 4 to both sides: x² + 4x + 4 = 9. The left side is (x + 2)² = 9. Taking square roots, x + 2 = ±3. So x = −2 + 3 = 1 or x = −2 − 3 = −5.

Numericals with worked steps

Solve 2x² + 7x + 3 = 0 by factorisation.

Find two numbers with product 2 × 3 = 6 and sum 7. They are 6 and 1.
2x² + 6x + x + 3 = 0.
2x(x + 3) + 1(x + 3) = 0.
(2x + 1)(x + 3) = 0.
2x + 1 = 0 gives x = −1/2, and x + 3 = 0 gives x = −3.

Answer: x = −1/2 or x = −3

Solve x² + 6x − 7 = 0 by completing the square.

x² + 6x = 7.
Add (6/2)² = 9 to both sides: x² + 6x + 9 = 16.
(x + 3)² = 16, so x + 3 = ±4.
x = −3 + 4 = 1 or x = −3 − 4 = −7.

Answer: x = 1 or x = −7

Solve 3x² − 5x − 2 = 0 by the quadratic formula.

a = 3, b = −5, c = −2.
b² − 4ac = 25 − 4 × 3 × (−2) = 25 + 24 = 49.
x = (5 ± √49) / 6 = (5 ± 7) / 6.
x = 12/6 = 2 or x = −2/6 = −1/3.

Answer: x = 2 or x = −1/3

The product of two consecutive positive integers is 56. Find them.

Let the integers be x and x + 1, so x(x + 1) = 56.
x² + x − 56 = 0.
(x + 8)(x − 7) = 0, so x = −8 or x = 7.
The integers are positive, so reject x = −8.
The integers are 7 and 8. Check: 7 × 8 = 56.

Answer: 7 and 8

Solve x⁴ − 13x² + 36 = 0.

Let y = x². Then y² − 13y + 36 = 0.
(y − 4)(y − 9) = 0, so y = 4 or y = 9.
x² = 4 gives x = ±2, and x² = 9 gives x = ±3.

Answer: x = ±2 or x = ±3

Common mistakes

Forgetting to move every term to one side before factorising. For (x − 2)(x − 3) = 6 you cannot say that each bracket equals 6.
Dropping the negative root when taking a square root. The square root gives ±.
Making a sign error for b in the formula. If the equation is x² − 5x + 6 = 0 then b = −5, so −b = 5.
Giving a root that does not fit a word problem, such as a negative length or a negative number of items.

Practise this chapter

Practice questions and timed test Printable worksheet