A-Level Quadratic functions Practice Questions

Free A-Level Quadratic functions practice questions with full step-by-step worked solutions. Covers completing the square, fractions, turning point, minimum value. Practise exam-style problems and check your method.

completing the squarefractionsturning pointminimum valuediscriminantrepeated root
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
Express x2+6x+5x^2+6x+5 in the form (x+a)2+b(x+a)^2+b.
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Worked solution

  1. Identify the number in front of x

    b=6,c=5b = 6,\quad c = 5

    To complete the square on x^2 + 6x + 5 we focus on the coefficient of x, which is 6. Completing the square rewrites the expression using a perfect square bracket.

  2. Halve the coefficient of x

    62=3\frac{6}{2} = 3

    Halving the coefficient of x gives the number that goes inside the bracket. This works because \left(x + 3\right)^2 expands to give x^2 + 6x plus an extra piece.

  3. Write the squared bracket

    (x+3)2=x2+6x+9\left(x + 3\right)^2 = x^2 + 6x + 9

    Squaring the bracket recreates the x^2 and 6x terms, but it also adds an unwanted 9. We will cancel that extra number in the next step.

  4. Subtract the extra number and add c

    x2+6x+5=(x+3)2−9+5x^2 + 6x + 5 = \left(x + 3\right)^2 - 9 + 5

    We take away the extra 9 that the bracket introduced, then bring down the original constant 5. This keeps the expression exactly equal to what we started with.

  5. Simplify the constant

    (x+3)2−4\left(x + 3\right)^2 - 4

    Combining -9 and 5 gives -4. The expression is now in completed-square form ax^2 style: \left(x + 3\right)^2 - 4.

Answer
(x+3)2−4\left(x + 3\right)^2 - 4
Question 2
3 markseasy
Sketch the graph of y=x2−4y=x^2-4, showing where it crosses the axes.
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Worked solution

  1. Decide the shape

    a=1>0⇒U-shapeda = 1 >0 \Rightarrow \text{U-shaped}

    The sign of the x^2 coefficient decides the shape: positive gives a U-shape (minimum), negative gives an n-shape (maximum). Here a = 1.

  2. Find the y-intercept

    x=0⇒y=−4x = 0 \Rightarrow y = -4

    Putting x = 0 gives the point where the curve crosses the y-axis, which is (0, -4). This is always just the constant term.

  3. Find the x-intercepts

    x2−4=0⇒x=−2, 2x^2 - 4 = 0 \Rightarrow x = -2,\ 2

    Setting y = 0 and factorising gives where the curve crosses the x-axis: x = -2 and x = 2.

  4. Find the turning point

    vertex=(0, −4)\text{vertex} = \left(0,\ -4\right)

    Completing the square (or using x = -b/2a) gives the turning point at (0, -4). This is the lowest or highest point of the curve.

  5. Draw the sketch

    plot the key features and join with a smooth curve\text{plot the key features and join with a smooth curve}

    We mark the intercepts and turning point, then draw a smooth symmetrical parabola through them. Choosing the correct graph is just matching these features.

Answer
See graph option A (correct sketch).
Question 3
3 marksintermediate
Find the values of kk for which 2x2+kx+8=02x^2+kx+8=0 has equal roots.
Show worked solution

Worked solution

  1. State the condition for equal (repeated) roots

    Δ=b2−4ac=0\Delta = b^2 - 4ac = 0

    For equal (repeated) roots the discriminant b^2-4ac must be zero. We will build the discriminant in terms of k and apply this condition.

  2. Identify a, b and c

    a=2,b=k,c=8a = 2,\quad b = k,\quad c = 8

    Read the coefficients from the quadratic, keeping the unknown k where it appears. Take care with any negative signs.

  3. Form the discriminant

    Δ=(k)2−4(2)(8)\Delta = \left(k\right)^2 - 4\left(2\right)\left(8\right)

    Substitute the coefficients into b^2-4ac. Brackets keep everything correct when we expand.

  4. Expand and simplify

    Δ=k2−64\Delta = k^{2} - 64

    Expanding and collecting like terms gives the discriminant as k^{2} - 64. This is now an expression purely in k.

  5. Apply the condition

    k2−64=0k^{2} - 64 = 0

    We now impose the condition by setting the discriminant equal to zero.

  6. Factorise / solve the quadratic in k

    (k−8)(k+8)=0\left(k - 8\right) \left(k + 8\right) = 0

    Factorising gives a quadratic in k. Setting each factor to zero yields the required values.

  7. Solve for k

    k=−8ork=8k = -8 \quad\text{or}\quad k = 8

    These are the value(s) of k that make the discriminant zero, giving the repeated root.

Answer
k=−8 or k=8k = -8 \text{ or } k = 8
Question 4
5 markshard
Find the values of kk for which x2+(k−3)x+k=0x^2+(k-3)x+k=0 has equal roots.
Show worked solution

Worked solution

  1. State the condition for equal (repeated) roots

    Δ=b2−4ac=0\Delta = b^2 - 4ac = 0

    For equal (repeated) roots the discriminant b^2-4ac must be zero. We will build the discriminant in terms of k and apply this condition.

  2. Identify a, b and c

    a=1,b=k−3,c=ka = 1,\quad b = k - 3,\quad c = k

    Read the coefficients from the quadratic, keeping the unknown k where it appears. Take care with any negative signs.

  3. Form the discriminant

    Δ=(k−3)2−4(1)(k)\Delta = \left(k - 3\right)^2 - 4\left(1\right)\left(k\right)

    Substitute the coefficients into b^2-4ac. Brackets keep everything correct when we expand.

  4. Expand and simplify

    Δ=k2−10k+9\Delta = k^{2} - 10 k + 9

    Expanding and collecting like terms gives the discriminant as k^{2} - 10 k + 9. This is now an expression purely in k.

  5. Apply the condition

    k2−10k+9=0k^{2} - 10 k + 9 = 0

    We now impose the condition by setting the discriminant equal to zero.

  6. Factorise / solve the quadratic in k

    (k−9)(k−1)=0\left(k - 9\right) \left(k - 1\right) = 0

    Factorising gives a quadratic in k. Setting each factor to zero yields the required values.

  7. Solve for k

    k=1ork=9k = 1 \quad\text{or}\quad k = 9

    These are the value(s) of k that make the discriminant zero, giving the repeated root.

  8. Recall the overall strategy

    identify→apply the method→check\text{identify} \to \text{apply the method} \to \text{check}

    It helps to see the shape of the whole method: read off the numbers, apply the right quadratic technique, then check the result. Keeping this plan in mind stops you getting lost mid-question.

  9. Interpret the answer in words

    k=1 or k=9k = 1 \text{ or } k = 9

    Before writing the final line, say out loud what the answer actually means for the curve or equation. Understanding it, rather than just copying symbols, makes it stick.

  10. Explain why the method works

    Δ=b2−4ac,a(x+p)2+q\Delta = b^2-4ac,\quad a(x+p)^2+q

    This question relies on two big ideas: the discriminant counts the real roots, and completed-square form reveals the turning point. Knowing why each tool works lets you choose it confidently.

Answer
k=1 or k=9k = 1 \text{ or } k = 9
Question 5
6 markschallenging
Solve (x2−2x)2−11(x2−2x)+24=0(x^2-2x)^2-11(x^2-2x)+24=0.
Show worked solution

Worked solution

  1. Spot the repeated bracket

    (x2−2x)2−11(x2−2x)+24=0\left(x^{2} - 2 x\right)^2 - 11\left(x^{2} - 2 x\right) + 24 = 0

    The expression (x2−2x)(x^2-2x) appears squared and on its own, so the whole equation is a quadratic in that bracket. Treating the bracket as a single object is the trick here.

  2. Substitute for the bracket

    let u=x2−2x\text{let } u = x^{2} - 2 x

    We let u stand for the repeated bracket. The equation immediately becomes a simple quadratic in u.

  3. Solve the quadratic in u

    (u−8)(u−3)=0⇒u=3 or u=8\left(u - 8\right) \left(u - 3\right) = 0 \Rightarrow u = 3 \text{ or } u = 8

    Factorising and setting each bracket to zero gives the values of u.

  4. Return to x for each value

    x2−2x=3⇒x=−1, 3x2−2x=8⇒x=−2, 4x^{2} - 2 x = 3 \Rightarrow x = -1,\ 3 \quad x^{2} - 2 x = 8 \Rightarrow x = -2,\ 4

    We replace u by the bracket again and solve each resulting equation for x. Each value of u can give one or two x-values.

  5. List all solutions

    x=−2, −1, 3, 4x = -2,\ -1,\ 3,\ 4

    Collecting the results gives all 4 solutions. Substituting any of them back into the original equation confirms they work.

  6. Recall the overall strategy

    identify→apply the method→check\text{identify} \to \text{apply the method} \to \text{check}

    It helps to see the shape of the whole method: read off the numbers, apply the right quadratic technique, then check the result. Keeping this plan in mind stops you getting lost mid-question.

  7. Interpret the answer in words

    x=−2, −1, 3, 4x = -2,\ -1,\ 3,\ 4

    Before writing the final line, say out loud what the answer actually means for the curve or equation. Understanding it, rather than just copying symbols, makes it stick.

  8. Explain why the method works

    Δ=b2−4ac,a(x+p)2+q\Delta = b^2-4ac,\quad a(x+p)^2+q

    This question relies on two big ideas: the discriminant counts the real roots, and completed-square form reveals the turning point. Knowing why each tool works lets you choose it confidently.

  9. Connect to completing the square

    ax2+bx+c=a(x+b2a)2+(c−b24a)ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)

    Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.

  10. Watch out for a common error

    keep signs and the ± carefully\text{keep signs and the } \pm \text{ carefully}

    The most common slips here are dropping a minus sign when substituting and forgetting the plus-or-minus when square-rooting. Slowing down at those points saves easy marks.

  11. Relate the result to the graph

    roots, y-intercept and turning point\text{roots, } y\text{-intercept and turning point}

    Every algebraic answer corresponds to a feature of the parabola: where it crosses the axes, its lowest or highest point, or how many times it meets a line. Picturing the graph is a strong final check.

  12. State the general principle

    the discriminant and vertex form describe every parabola\text{the discriminant and vertex form describe every parabola}

    The same ideas of discriminant (number of roots) and completed-square form (turning point) apply to all quadratics. Recognising the pattern lets you tackle unfamiliar questions.

  13. Double-check the reasoning

    re-read the question and confirm each line follows\text{re-read the question and confirm each line follows}

    Read the question once more and make sure every line follows logically from the one before. A quick review catches careless mistakes before they cost marks.

  14. Check the answer makes sense

    does the answer fit the situation?\text{does the answer fit the situation?}

    Ask whether the answer is sensible: lengths and areas must be positive, and the number of solutions should match the graph. A quick reasonableness check is the mark of a careful mathematician.

  15. Summarise the result

    x=−2, −1, 3, 4x = -2,\ -1,\ 3,\ 4

    Pulling the working together into one clear statement makes the solution easy to follow and easy to mark. This is a good habit for every multi-step question.

Answer
x=−2, −1, 3, 4x = -2,\ -1,\ 3,\ 4

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