Class 10 Quadratic Equations: Practice and Worked Solutions

Worked example: turn an area into an equation

A rectangular garden has area 48 m². Its length is 2 m more than its width. Find its dimensions. Let the width be x m, so the length is (x + 2) m. Area gives x(x + 2) = 48, hence x² + 2x − 48 = 0.

Factorise: (x + 8)(x − 6) = 0, so x = −8 or x = 6. A width cannot be negative, so the width is 6 m and the length is 8 m. Check both conditions: 6 × 8 = 48 and 8 − 6 = 2.

The useful habit is to keep the meaning of x beside the algebra. If you stop at the two roots, you have not finished answering the measurement question. If your equation used x + 2 as the width, your final interpretation must follow that definition instead.

Choose a method before calculating

• Put every term on one side first, then identify a, b and c in ax² + bx + c = 0. Check that a is non-zero; otherwise the equation is not quadratic.
• Try factorisation when the factors are straightforward. For 2x² − 7x + 3, split −7x as −6x − x and group the terms.
• Use x = (−b ± √(b² − 4ac))/(2a) when factors are not obvious. Substitute signed values in brackets so a negative b does not create a sign error.
• If the question asks only about the nature of the real roots, calculate D = b² − 4ac. D > 0 gives two distinct real roots; D = 0 gives equal real roots; D < 0 gives no real roots.
• For a word problem, define the variable and its restrictions, form the equation, solve and then test each result against the original conditions.

Original practice: attempt before checking

Use paper and pencil. For a short first session, attempt Q1, Q3 and Q5. Cover or scroll past the solution section until you have written a complete attempt. No marks are assigned: the purpose is to locate the step that needs work.

• Q1. Solve x² − 7x + 12 = 0 by factorisation. Verify both roots by substitution.
• Q2. Solve 2x² − 7x + 3 = 0. Show the middle-term split or another valid method.
• Q3. Without finding the roots, determine their nature for 3x² − 4x + 2 = 0.
• Q4. Find all values of k for which x² + kx + 9 = 0 has equal real roots.
• Q5. The product of two consecutive positive integers is 72. Form an equation and find the integers.
• Q6. Solve x² − 2x − 1 = 0 using the quadratic formula. Give exact answers.
• Q7. Case study: a rectangular display board has area 60 cm². Its length is 4 cm more than its width. (a) Define x and form the quadratic equation. (b) Find both algebraic roots. (c) Choose the valid dimensions and explain why the other root is rejected.
• Q8. A student solves x² − 5x = 0 by dividing both sides by x and reports only x = 5. Explain the missing solution and solve without dividing by x.

Solutions: compare the reasoning, not only the answer

Q1. x² − 7x + 12 = (x − 3)(x − 4). The roots are 3 and 4. Substitution gives 9 − 21 + 12 = 0 and 16 − 28 + 12 = 0. A frequent error is choosing factors with the right product but the wrong sum.

Q2. 2x² − 6x − x + 3 = 0 becomes 2x(x − 3) − (x − 3) = 0, so (2x − 1)(x − 3) = 0. Hence x = 1/2 or 3. Multiplying the factors back out is a quick check on the grouping sign.

Q3. a = 3, b = −4 and c = 2. D = (−4)² − 4(3)(2) = 16 − 24 = −8. There are no real roots. A negative discriminant does not mean that the equation has a negative real root.

Q4. Equal real roots require D = 0: k² − 4(1)(9) = 0, so k² = 36 and k = 6 or −6. For k = 6 the equation is (x + 3)² = 0; for k = −6 it is (x − 3)² = 0. Both parameter values are valid.

Q5. Let the smaller integer be n; the next is n + 1. n(n + 1) = 72 gives n² + n − 72 = (n + 9)(n − 8) = 0. The algebraic choices are −9 and 8, but the integers must be positive. They are 8 and 9; 8 × 9 = 72.

Q6. a = 1, b = −2, c = −1. D = 4 + 4 = 8. x = (2 ± √8)/2 = 1 ± √2. Keep the square root exact; substituting a rounded decimal too early can obscure the check.

Q7. Let width = x cm, length = x + 4 cm. x(x + 4) = 60 gives x² + 4x − 60 = (x + 10)(x − 6) = 0. The roots are −10 and 6. Width is 6 cm and length is 10 cm; the negative root cannot be a width. Check 6 × 10 = 60 and 10 − 6 = 4.

Q8. Dividing by x assumes x ≠ 0, which removes a possible solution. Instead write x(x − 5) = 0, giving x = 0 or 5. Substitute both in the original equation. Cancellation is valid only after checking that the cancelled quantity cannot be zero.

Match the error to a correction task

• Wrong coefficients or signs: rewrite Q2 and Q6 in standard form and circle the signed values of a, b and c before choosing a method.
• Cannot form the equation: draw the rectangle in Q7 and write width, length and area in separate lines. Explain why multiplication, rather than addition, represents the area.
• Only one root or an invalid measurement: revisit Q4, Q5 and Q8. List every algebraic result before applying the situation's restrictions.
• Correct answer but unexplained work: write the factorisation or formula substitution, both roots and the contextual check. Compare the format with the official marking scheme when you return to an official paper.

Check the correction on changed questions

Retest after a break: R1. Solve 3x² − 10x + 3 = 0. R2. A rectangle has area 84 m² and length 5 m more than its width. Find the dimensions. Try both without copying the earlier working.

Retest checks: R1 factorises as (3x − 1)(x − 3), giving 1/3 and 3. R2 gives x² + 5x − 84 = (x + 12)(x − 7), so width 7 m and length 12 m. If the same mistake returns, work on that step before starting a full paper.

For a parent or tutor, show the first attempt, one corrected line and the independent retest. That is more useful than reporting only how many answers matched. Successful work on these questions is evidence about these skills, not a predicted board score.

Move from chapter correction to a paper

When you can explain the setup and check the roots unaided, return to the appropriate official paper. Use its marking scheme for that paper's requirements. For a different algebra skill, try the AP practice: its main challenge is deciding whether the question asks for a single term or a total.

Read the explanation behind this skill