Hard A-Level Quadratic functions Questions

Challenging, exam-style A-Level Quadratic functions questions with worked solutions. Stretch yourself on the hardest completing the square, maximum value, discriminant, distinct roots problems.

completing the squaremaximum valuediscriminantdistinct rootstangentno real roots

A-Level34 questionsStep-by-step solutions

Question 1

6 markschallenging

Solve

(x^2-2x)^2-11(x^2-2x)+24=0

Show worked solution

Worked solution

Spot the repeated bracket
$\left(x^{2} - 2 x\right)^2 - 11\left(x^{2} - 2 x\right) + 24 = 0$
The expression $(x^2-2x)$ 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.
Substitute for the bracket
$\text{let } u = x^{2} - 2 x$
We let u stand for the repeated bracket. The equation immediately becomes a simple quadratic in u.
Solve the quadratic in u
$\left(u - 8\right) \left(u - 3\right) = 0 \Rightarrow u = 3 \text{ or } u = 8$
Factorising and setting each bracket to zero gives the values of u.
Return to x for each value
$x^{2} - 2 x = 3 \Rightarrow x = -1,\ 3 \quad x^{2} - 2 x = 8 \Rightarrow x = -2,\ 4$
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.
List all solutions
$x = -2,\ -1,\ 3,\ 4$
Collecting the results gives all 4 solutions. Substituting any of them back into the original equation confirms they work.
Recall the overall strategy
$\text{identify} \to \text{apply the method} \to \text{check}$
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.
Interpret the answer in words
$x = -2,\ -1,\ 3,\ 4$
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.
Explain why the method works
$\Delta = b^2-4ac,\quad a(x+p)^2+q$
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.
Connect to completing the square
$ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$
Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.
Watch out for a common error
$\text{keep signs and the } \pm \text{ carefully}$
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.
Relate the result to the graph
$\text{roots, } y\text{-intercept and turning point}$
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.
State the general principle
$\text{the discriminant and vertex form describe every parabola}$
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.
Double-check the reasoning
$\text{re-read the question and confirm each line follows}$
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.
Check the answer makes sense
$\text{does the answer fit the situation?}$
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.
Summarise the result
$x = -2,\ -1,\ 3,\ 4$
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,\ 4

Question 2

6 markschallenging

The line

y=2x+c

is a tangent to the curve

y=x^2-3x+7

. Find the value of

c

Show worked solution

Worked solution

Set the two expressions equal to form one equation
$x^2-3x+7 = 2x+c$
A tangent touches the curve where their y-values are equal, so we set the expressions equal.
Rearrange into a single quadratic equal to zero
$x^2-5x+(7-c) = 0$
Collecting terms gives a quadratic; a tangent touches once, so it has equal roots.
State the condition for equal (repeated) roots
$\Delta = b^2 - 4ac = 0$
For equal (repeated) roots the discriminant b^2-4ac must be zero. We will build the discriminant in terms of c and apply this condition.
Identify a, b and c
$a = 1,\quad b = -5,\quad c = 7 - c$
Read the coefficients from the quadratic, keeping the unknown c where it appears. Take care with any negative signs.
Form the discriminant
$\Delta = \left(-5\right)^2 - 4\left(1\right)\left(7 - c\right)$
Substitute the coefficients into b^2-4ac. Brackets keep everything correct when we expand.
Expand and simplify
$\Delta = 4 c - 3$
Expanding and collecting like terms gives the discriminant as 4 c - 3. This is now an expression purely in c.
Apply the condition
$4 c - 3 = 0$
We now impose the condition by setting the discriminant equal to zero.
Solve for c
$c = \frac{3}{4}$
These are the value(s) of c that make the discriminant zero, giving the repeated root.
Recall the overall strategy
$\text{identify} \to \text{apply the method} \to \text{check}$
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.
Interpret the answer in words
$c = \frac{3}{4}$
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.
Explain why the method works
$\Delta = b^2-4ac,\quad a(x+p)^2+q$
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.
Connect to completing the square
$ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$
Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.
Watch out for a common error
$\text{keep signs and the } \pm \text{ carefully}$
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.
Relate the result to the graph
$\text{roots, } y\text{-intercept and turning point}$
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.
State the general principle
$\text{the discriminant and vertex form describe every parabola}$
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.

Answer

c = \frac{3}{4}

Question 3

6 markschallenging

Find the range of values of

k

for which the line

y=kx+2

does not meet the curve

y=x^2+3x+6

Show worked solution

Worked solution

Set the two expressions equal to form one equation
$x^2+3x+6 = kx+2$
The line meets the curve where their y-values are equal. If they never meet, this equation has no real solutions.
Rearrange into a single quadratic equal to zero
$x^2+(3-k)x+4 = 0$
Collecting terms gives a quadratic; 'does not meet' means this quadratic has no real roots.
State the condition for no real roots
$\Delta = b^2 - 4ac < 0$
For no real roots the discriminant b^2-4ac must be negative. We will build the discriminant in terms of k and apply this condition.
Identify a, b and c
$a = 1,\quad b = 3 - k,\quad c = 4$
Read the coefficients from the quadratic, keeping the unknown k where it appears. Take care with any negative signs.
Form the discriminant
$\Delta = \left(3 - k\right)^2 - 4\left(1\right)\left(4\right)$
Substitute the coefficients into b^2-4ac. Brackets keep everything correct when we expand.
Expand and simplify
$\Delta = k^{2} - 6 k - 7$
Expanding and collecting like terms gives the discriminant as k^{2} - 6 k - 7. This is now an expression purely in k.
Apply the condition
$k^{2} - 6 k - 7 < 0$
We now impose the condition by setting the discriminant less than zero.
Factorise the quadratic in k
$\left(k - 7\right) \left(k + 1\right) < 0$
Factorising helps us find the critical values k = -1 and k = 7, where the discriminant changes sign.
Read off the solution set
$-1 < k < 7$
A sketch of the upward parabola in k shows where it is below the axis. That gives the range of k shown.
Recall the overall strategy
$\text{identify} \to \text{apply the method} \to \text{check}$
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.
Interpret the answer in words
$-1 < k < 7$
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.
Explain why the method works
$\Delta = b^2-4ac,\quad a(x+p)^2+q$
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.
Connect to completing the square
$ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$
Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.
Watch out for a common error
$\text{keep signs and the } \pm \text{ carefully}$
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.
Relate the result to the graph
$\text{roots, } y\text{-intercept and turning point}$
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.

Answer

-1 < k < 7

Question 4

6 markschallenging

Solve

x^2-x-1=0

, giving your answers in exact (surd) form.

Show worked solution

Worked solution

Recognise it does not factorise easily
$x^2 - x - 1 = 0$
There is no simple pair of whole numbers that factorises this quadratic, so we use the quadratic formula. The formula works for every quadratic.
Write the quadratic formula
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
This formula comes from completing the square on the general quadratic. Learning it means you can always solve, even when factorising fails.
Identify a, b and c
$a = 1,\quad b = -1,\quad c = -1$
Read the three coefficients carefully, keeping any negative signs. Getting these right is essential before substituting.
Substitute into the formula
$x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$
Put each value in using brackets, especially around negatives. This is the step where careful substitution prevents sign errors.
Work out the discriminant
$x = \dfrac{1 \pm \sqrt{5}}{2}$
The number under the square root is 5. Because it is positive there are two different real solutions.
Simplify the surd
$\sqrt{5} = \sqrt{5}$
We simplify \sqrt{5} by taking out the largest square factor, giving \sqrt{5}. Leaving surds exact is expected at A-Level unless a decimal is asked for.
Write the exact solutions
$x = \frac{1}{2} \pm \frac{1}{2}\sqrt{5}$
We simplify the fraction to give the two exact solutions. The \pm sign captures both the larger and smaller root at once.
Recall the overall strategy
$\text{identify} \to \text{apply the method} \to \text{check}$
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.
Interpret the answer in words
$x = \frac{1}{2} \pm \frac{1}{2}\sqrt{5}$
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.
Explain why the method works
$\Delta = b^2-4ac,\quad a(x+p)^2+q$
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.
Connect to completing the square
$ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$
Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.
Watch out for a common error
$\text{keep signs and the } \pm \text{ carefully}$
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.
Relate the result to the graph
$\text{roots, } y\text{-intercept and turning point}$
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.
State the general principle
$\text{the discriminant and vertex form describe every parabola}$
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.
Double-check the reasoning
$\text{re-read the question and confirm each line follows}$
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.

Answer

x = \frac{1}{2} \pm \frac{1}{2}\sqrt{5}

Question 5

6 markschallenging

Express

4x^2-4x+5

in completed-square form and hence find its minimum value.

Show worked solution

Worked solution

Factor the leading coefficient from the first two terms
$4x^2 - 4x + 5 = 4\left(x^2 + -1x\right) + 5$
When the number in front of x^2 is not 1 we first take it out as a common factor of the x^2 and x terms. The constant 5 is left outside the bracket for now.
Halve the coefficient of x inside the bracket
$\frac{-1}{2} = -\frac{1}{2}$
Inside the bracket the coefficient of x is -1. Halving it gives -\frac{1}{2}, the number for the inner squared bracket.
Complete the square inside the bracket
$x^2 + -1x = \left(x + -\frac{1}{2}\right)^2 - \frac{1}{4}$
We rewrite the inside as a perfect square and subtract the extra \frac{1}{4} it creates. This is the same completing-the-square idea, just done inside the bracket.
Multiply the factor back in
$4\left[\left(x + -\frac{1}{2}\right)^2 - \frac{1}{4}\right] + 5$
Now we put the factor 4 back so nothing changes. Everything inside the square brackets is multiplied by 4.
Expand the constant term
$4\left(x + -\frac{1}{2}\right)^2 - 1 + 5$
Multiplying 4 by -\frac{1}{4} gives -1. Only the constant is affected; the squared bracket keeps its factor of 4.
Simplify the constant
$4\left(x - \frac{1}{2}\right)^2 + 4$
Combining -1 and 5 gives 4. The completed-square form is 4\left(x - \frac{1}{2}\right)^2 + 4.
Read off the minimum value
$min\;y = 4 \text{ at } x = \frac{1}{2}$
In completed-square form the constant q is the minimum value of the expression, reached when the squared bracket is zero (at x = \frac{1}{2}). The bracket can never be negative, so y cannot go past 4.
Confirm with a sketch
$\text{plot the curve and read off the key features}$
Drawing the parabola lets us see that the answer agrees with the graph. A quick sketch is a powerful way to catch mistakes.
Recall the overall strategy
$\text{identify} \to \text{apply the method} \to \text{check}$
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.
Interpret the answer in words
$4$
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.
Explain why the method works
$\Delta = b^2-4ac,\quad a(x+p)^2+q$
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.
Connect to completing the square
$ax^2+bx+c = a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$
Completing the square underpins the quadratic formula, the turning point and the range. Seeing the link back to it ties the whole topic together.
Watch out for a common error
$\text{keep signs and the } \pm \text{ carefully}$
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.
Relate the result to the graph
$\text{roots, } y\text{-intercept and turning point}$
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.
State the general principle
$\text{the discriminant and vertex form describe every parabola}$
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.

Answer

4

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