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رياضيات GCSE

Quadratic Equations Practice Questions with Worked Solutions

Twelve quadratics with the route chosen for each: factorising, completing the square or the formula, plus one where a negative discriminant ends the question.

الإجابة باختصار

Twelve quadratics solved in full, with the method chosen to fit each one rather than one method used throughout. Seven factorise, one needs completing the square, two need the formula, one has a negative discriminant and no real solutions, and the last asks for the completed square form.

الأسئلة

  1. سؤال 1أساسي3 درجات

    Solve x² - 5x + 6 = 0

    أظهر الحل

    x = 2 or x = 3

    1. Find two numbers multiplying to +6 and adding to -5: they are -2 and -3.
    2. (x - 2)(x - 3) = 0
    3. A product is zero only when one of the factors is zero.
    4. x = 2 or x = 3
  2. سؤال 2أساسي3 درجات

    Solve x² + 7x + 12 = 0

    أظهر الحل

    x = -3 or x = -4

    1. Two numbers multiplying to +12 and adding to +7: 3 and 4.
    2. (x + 3)(x + 4) = 0
    3. Set each bracket to zero: x + 3 = 0 or x + 4 = 0.
    4. x = -3 or x = -4. The signs in the brackets are positive; the solutions are negative.
  3. سؤال 3أساسي2 درجات

    Solve x² - 9 = 0

    أظهر الحل

    x = 3 or x = -3

    1. There is no x term, and 9 is a square, so this is a difference of two squares: x² - 3².
    2. (x - 3)(x + 3) = 0
    3. x = 3 or x = -3
    4. Square-rooting both sides gives the same thing, provided both signs are kept: x = ±3.
  4. سؤال 4أساسي2 درجات

    Solve x² + 2x = 0

    أظهر الحل

    x = 0 or x = -2

    1. There is no constant term, so x is a common factor of both terms.
    2. x(x + 2) = 0
    3. Either x = 0 or x + 2 = 0.
    4. x = 0 or x = -2. Dividing both sides by x would throw away the root x = 0.
  5. سؤال 5جوهري3 درجات

    Solve x² + 3x - 10 = 0

    أظهر الحل

    x = -5 or x = 2

    1. The constant is negative, so the two numbers have opposite signs.
    2. They multiply to -10 and add to +3: +5 and -2.
    3. (x + 5)(x - 2) = 0
    4. x = -5 or x = 2
  6. سؤال 6جوهري3 درجات

    Solve 2x² + 7x + 3 = 0

    أظهر الحل

    x = -1/2 or x = -3

    1. Here a is 2, so look for two numbers multiplying to 2 × 3 = 6 and adding to 7: 6 and 1.
    2. Split the middle term: 2x² + 6x + x + 3 = 0.
    3. Factor in pairs: 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0.
    4. 2x + 1 = 0 gives x = -1/2; x + 3 = 0 gives x = -3.
  7. سؤال 7جوهري4 درجات

    Solve 6x² - x - 2 = 0

    أظهر الحل

    x = 2/3 or x = -1/2

    1. a × c = 6 × -2 = -12. Two numbers multiplying to -12 and adding to -1: -4 and +3.
    2. Split the middle term: 6x² - 4x + 3x - 2 = 0.
    3. Factor in pairs: 2x(3x - 2) + 1(3x - 2) = 0, so (3x - 2)(2x + 1) = 0.
    4. 3x - 2 = 0 gives x = 2/3; 2x + 1 = 0 gives x = -1/2.
  8. سؤال 8جوهري4 درجات

    Solve x² + 6x + 4 = 0 by completing the square. Give exact answers and then answers to 3 significant figures.

    أظهر الحل

    x = -3 + √5 or x = -3 - √5, which are -0.764 and -5.24 to 3 s.f.

    1. No whole numbers multiply to 4 and add to 6, so it does not factorise.
    2. Halve the coefficient of x: (x + 3)² = x² + 6x + 9, which is 5 more than x² + 6x + 4.
    3. So x² + 6x + 4 = (x + 3)² - 9 + 4 = (x + 3)² - 5 = 0.
    4. (x + 3)² = 5, so x + 3 = ±√5.
    5. x = -3 + √5 = -0.764 or x = -3 - √5 = -5.24 (3 s.f.).
  9. سؤال 9جوهري3 درجات

    Solve 3x² - 7x + 2 = 0 using the quadratic formula.

    أظهر الحل

    x = 2 or x = 1/3

    1. a = 3, b = -7, c = 2.
    2. b² - 4ac = (-7)² - 4 × 3 × 2 = 49 - 24 = 25, and √25 = 5.
    3. x = (7 ± 5) / (2 × 3) = (7 ± 5) / 6
    4. x = 12/6 = 2 or x = 2/6 = 1/3.
    5. The discriminant was a perfect square, which is why this one also factorises as (3x - 1)(x - 2) = 0.
  10. سؤال 10متقدّم4 درجات

    Solve 2x² - 4x - 3 = 0. Give exact answers in surd form and answers to 2 decimal places.

    أظهر الحل

    x = (2 + √10)/2 or x = (2 - √10)/2, which are 2.58 and -0.58 to 2 d.p.

    1. a = 2, b = -4, c = -3.
    2. b² - 4ac = 16 - 4 × 2 × -3 = 16 + 24 = 40, which is not a perfect square, so it will not factorise.
    3. x = (4 ± √40) / 4, and √40 = √4 × √10 = 2√10.
    4. x = (4 ± 2√10)/4, and dividing every term by 2 gives x = (2 ± √10)/2.
    5. √10 = 3.1623, so x = 2.58 or x = -0.58 to 2 d.p.
  11. سؤال 11متقدّم3 درجات

    Solve x² + 4x + 7 = 0

    أظهر الحل

    No real solutions

    1. a = 1, b = 4, c = 7.
    2. b² - 4ac = 16 - 4 × 1 × 7 = 16 - 28 = -12.
    3. The discriminant is negative, and no real number squares to give a negative.
    4. So x² + 4x + 7 = 0 has no real solutions. Completing the square shows why: (x + 2)² + 3 is never less than 3.
  12. سؤال 12متقدّم4 درجات

    Write x² - 8x + 3 in the form (x + a)² + b, and state the minimum value of the expression and the value of x at which it occurs.

    أظهر الحل

    (x - 4)² - 13; minimum value -13, at x = 4

    1. Halve the coefficient of x: half of -8 is -4, so start from (x - 4)².
    2. (x - 4)² = x² - 8x + 16, which is 16 more than x² - 8x.
    3. So x² - 8x + 3 = (x - 4)² - 16 + 3 = (x - 4)² - 13.
    4. A square is never negative, so the smallest (x - 4)² can be is 0.
    5. The minimum value is -13, and it happens when x - 4 = 0, that is at x = 4.

أين يقع الخطأ عادةً

  • Dividing x² + 2x = 0 through by x. It looks like a simplification and it silently deletes the solution x = 0.
  • Reading the solutions straight off the brackets: (x + 3)(x + 4) = 0 gives x = -3 and x = -4, not +3 and +4.
  • Substituting b rather than -b into the formula, or evaluating (-7)² as -49 instead of 49.
  • Treating x = -3 ± √5 as a single answer and giving only one value, when the ± sign is holding two solutions.
  • Stopping when the discriminant comes out negative, as though the question had gone wrong. No real solutions is the answer and it earns the marks.

Try factorising, and know when to stop trying

Factorising is much the fastest route when it works, and seven of these twelve factorise. The problem is the time spent hunting for factors that are not there. x² + 6x + 4 has no pair of whole numbers multiplying to 4 and adding to 6, and a student can lose four minutes discovering that.

Thirty seconds spent on the discriminant settles it first. b² - 4ac for x² + 6x + 4 is 36 - 16 = 20, which is not a perfect square, so no whole-number factorisation exists and the search can be abandoned before it starts. For 3x² - 7x + 2 the discriminant is 25, a perfect square, and the factors are worth looking for.

One equation, three routes

3x² - 7x + 2 = 0 can be solved all three ways, and comparing them shows what each method costs. Factorising means splitting the middle term using 6 and 1, since 3 × 2 = 6: 3x² - 6x - x + 2 = 0 becomes 3x(x - 2) - 1(x - 2) = 0, so (3x - 1)(x - 2) = 0 and x = 1/3 or x = 2. Four lines.

The formula takes three lines and no thinking: b² - 4ac = 49 - 24 = 25, so x = (7 ± 5)/6, giving 2 and 1/3. Completing the square takes seven, because the leading 3 has to come out first: 3(x - 7/6)² - 25/12 = 0, so (x - 7/6)² = 25/36 and x = 7/6 ± 5/6.

The answers are identical. The choice is about time, and about which method you can carry out without a slip when the numbers are unfriendly. Completing the square earns its place elsewhere - it is the only one of the three that hands you the minimum point as well.

What b² - 4ac tells you before you solve anything

The discriminant is the part of the formula under the square root sign, and its sign decides how many real solutions exist. Positive means two, zero means one repeated root, negative means none.

That last case is a real answer, not a failure. x² + 4x + 7 = 0 has a discriminant of 16 - 28 = -12, and no real number squares to give a negative, so the equation has no real solutions. On a graph, the parabola never reaches the x-axis. Writing no real solutions is worth full marks; abandoning the question is worth nothing.

  • b² - 4ac > 0 - two distinct real solutions, the curve crosses the x-axis twice
  • b² - 4ac = 0 - one repeated solution, the curve touches the x-axis
  • b² - 4ac < 0 - no real solutions, the curve misses the x-axis
  • b² - 4ac a perfect square - the two solutions are rational, so it will factorise

أسئلة شائعة

Which method should I use on a quadratic?

Work out b² - 4ac first. If it is a perfect square, the equation factorises and that is the quickest route. If it is positive but not a perfect square, use the formula. Use completing the square when the question asks for it, or when you also need the minimum point or the line of symmetry.

Why does a quadratic usually have two answers?

Because the graph is a parabola, and a curve that comes down and goes back up crosses a horizontal line in two places. Solving f(x) = 0 asks where it crosses the x-axis. Two crossings give two solutions, a curve that just touches gives one, and a curve that misses gives none.

What does it mean if the discriminant is exactly zero?

One repeated solution. x² - 6x + 9 = 0 has b² - 4ac = 36 - 36 = 0, and it factorises to (x - 3)² = 0, giving x = 3 twice. Graphically the parabola touches the x-axis at a single point rather than cutting through it. Exam questions often phrase this as showing that a line is a tangent to a curve.

Should I give exact answers or decimals?

Whatever the question asks. If it says give your answers to 2 decimal places, round at the very end and not before. If it says leave your answer in surd form, then (2 + √10)/2 is the answer and 2.58 will lose marks. When neither is specified, give the exact form and add the decimal beside it.

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