Quadratic functions Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Quadratic functions questions. See exactly how to solve problems on completing the square, fractions, turning point, minimum value.

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.

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
Write x2−10x+2x^2-10x+2 in the form (x+a)2+b(x+a)^2+b.

Worked solution

  1. Identify the number in front of x

    b=−10,c=2b = -10,\quad c = 2

    To complete the square on x^2 - 10x + 2 we focus on the coefficient of x, which is -10. Completing the square rewrites the expression using a perfect square bracket.

  2. Halve the coefficient of x

    −102=−5\frac{-10}{2} = -5

    Halving the coefficient of x gives the number that goes inside the bracket. This works because \left(x + -5\right)^2 expands to give x^2 + -10x plus an extra piece.

  3. Write the squared bracket

    (x+−5)2=x2+−10x+25\left(x + -5\right)^2 = x^2 + -10x + 25

    Squaring the bracket recreates the x^2 and -10x terms, but it also adds an unwanted 25. We will cancel that extra number in the next step.

  4. Subtract the extra number and add c

    x2+−10x+2=(x+−5)2−25+2x^2 + -10x + 2 = \left(x + -5\right)^2 - 25 + 2

    We take away the extra 25 that the bracket introduced, then bring down the original constant 2. This keeps the expression exactly equal to what we started with.

  5. Simplify the constant

    (x−5)2−23\left(x - 5\right)^2 - 23

    Combining -25 and 2 gives -23. The expression is now in completed-square form ax^2 style: \left(x - 5\right)^2 - 23.

Answer
(x−5)2−23\left(x - 5\right)^2 - 23
Question 3
3 markseasy
Express x2+3x+1x^2+3x+1 in the form (x+a)2+b(x+a)^2+b.

Worked solution

  1. Identify the number in front of x

    b=3,c=1b = 3,\quad c = 1

    To complete the square on x^2 + 3x + 1 we focus on the coefficient of x, which is 3. Completing the square rewrites the expression using a perfect square bracket.

  2. Halve the coefficient of x

    32=32\frac{3}{2} = \frac{3}{2}

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

  3. Write the squared bracket

    (x+32)2=x2+3x+94\left(x + \frac{3}{2}\right)^2 = x^2 + 3x + \frac{9}{4}

    Squaring the bracket recreates the x^2 and 3x terms, but it also adds an unwanted \frac{9}{4}. We will cancel that extra number in the next step.

  4. Subtract the extra number and add c

    x2+3x+1=(x+32)2−94+1x^2 + 3x + 1 = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + 1

    We take away the extra \frac{9}{4} that the bracket introduced, then bring down the original constant 1. This keeps the expression exactly equal to what we started with.

  5. Simplify the constant

    (x+32)2−54\left(x + \frac{3}{2}\right)^2 - \frac{5}{4}

    Combining -\frac{9}{4} and 1 gives -\frac{5}{4}. The expression is now in completed-square form ax^2 style: \left(x + \frac{3}{2}\right)^2 - \frac{5}{4}.

Answer
(x+32)2−54\left(x + \frac{3}{2}\right)^2 - \frac{5}{4}
Question 4
3 markseasy
The curve y=x2−4x+7y=x^2-4x+7 has a minimum point. Find its coordinates.

Worked solution

  1. Identify the number in front of x

    b=−4,c=7b = -4,\quad c = 7

    To complete the square on x^2 - 4x + 7 we focus on the coefficient of x, which is -4. Completing the square rewrites the expression using a perfect square bracket.

  2. Halve the coefficient of x

    −42=−2\frac{-4}{2} = -2

    Halving the coefficient of x gives the number that goes inside the bracket. This works because \left(x + -2\right)^2 expands to give x^2 + -4x plus an extra piece.

  3. Write the squared bracket

    (x+−2)2=x2+−4x+4\left(x + -2\right)^2 = x^2 + -4x + 4

    Squaring the bracket recreates the x^2 and -4x terms, but it also adds an unwanted 4. We will cancel that extra number in the next step.

  4. Subtract the extra number and add c

    x2+−4x+7=(x+−2)2−4+7x^2 + -4x + 7 = \left(x + -2\right)^2 - 4 + 7

    We take away the extra 4 that the bracket introduced, then bring down the original constant 7. This keeps the expression exactly equal to what we started with.

  5. Simplify the constant

    (x−2)2+3\left(x - 2\right)^2 + 3

    Combining -4 and 7 gives 3. The expression is now in completed-square form ax^2 style: \left(x - 2\right)^2 + 3.

  6. Read off the turning point

    vertex=(2, 3)\text{vertex} = \left(2,\ 3\right)

    In the form a(x+p)^2+q the vertex is at (-p, q). Here that gives the turning point (2, 3); the graph shows this lowest/highest point.

Answer
(2, 3)\left(2,\ 3\right)
Question 5
2 markseasy
Find the minimum value of x2+8x+20x^2+8x+20.

Worked solution

  1. Find the x-coordinate of the turning point

    x=−b2a=−82(1)=−4x = -\dfrac{b}{2a} = -\dfrac{8}{2(1)} = -4

    The minimum of a quadratic is at its turning point, whose x-coordinate is x = -b/(2a). Substituting gives x = -4.

  2. Substitute to find the minimum value

    y=x2+8x+20∣x=−4=4y = x^2 + 8x + 20\big|_{x=-4} = 4

    Putting x = -4 back into the expression gives its minimum value. This is the y-coordinate of the turning point.

  3. State the minimum value

    min\;y = 4

    So the minimum value is 4. This links to completing the square, where the same constant appears as q.

Answer
44

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