Hard GCSE Completing square and formula Questions

Challenging, exam-style GCSE Completing square and formula questions with worked solutions. Stretch yourself on the hardest completing the square, coefficient of x squared not 1, quadratic formula, non-monic quadratic problems.

completing the squarecoefficient of x squared not 1quadratic formulanon-monic quadraticsurd formsimplifying surds
GCSE Higher34 questionsStep-by-step solutions
Question 1
5 markschallenging
A quadratic y=ax2+bx+cy = ax^2 + bx + c has a>0a > 0 and b2−4ac<0b^2 - 4ac < 0. Which statement about its graph and its completed square form is correct?
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Worked solution

  1. Interpret a>0a > 0

    a>0⇒U-shaped (opens upwards)a > 0 \Rightarrow \text{U-shaped (opens upwards)}

    A positive coefficient of x2x^2 gives a parabola with a minimum point.

  2. Interpret the discriminant

    b2−4ac<0⇒no real rootsb^2 - 4ac < 0 \Rightarrow \text{no real roots}

    No real roots means the curve never meets the xx-axis.

  3. Put the two together

    opens up, never crosses the axis\text{opens up, never crosses the axis}

    A U-shaped curve that never reaches the xx-axis must lie entirely above it.

  4. Write the completed square

    ax2+bx+c=a(x+b2a)2+qax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + q

    The general completed square form for any quadratic.

  5. Find qq in general

    q=c−b24a=4ac−b24aq = c - \frac{b^2}{4a} = \frac{4ac - b^2}{4a}

    Multiply out and compare constants.

  6. Relate qq to the discriminant

    q=−(b2−4ac)4aq = \frac{-(b^2 - 4ac)}{4a}

    4ac−b24ac - b^2 is exactly −(b2−4ac)-(b^2 - 4ac).

  7. Deduce the sign of qq

    b2−4ac<0 and a>0⇒q>0b^2 - 4ac < 0 \text{ and } a > 0 \Rightarrow q > 0

    A negative numerator negated becomes positive, and the denominator 4a4a is positive — so qq is positive.

  8. State the minimum value

    ymin=q>0y_{\text{min}} = q > 0

    The square term is never negative, so the smallest value yy takes is qq, which is above zero.

  9. Confirm no roots

    a(x+p)2+q=0⇒(x+p)2=−qa<0a(x + p)^2 + q = 0 \Rightarrow (x + p)^2 = -\frac{q}{a} < 0

    A square can never be negative, so there is no real solution — exactly as the discriminant predicted.

  10. Locate the turning point

    (−b2a, q)\left(-\frac{b}{2a},\ q\right)

    The minimum sits on the line of symmetry x=−b2ax = -\frac{b}{2a}.

  11. Note the yy-intercept

    y=c when x=0y = c \text{ when } x = 0

    Since the whole curve is above the axis, c>0c > 0 as well.

  12. Check with an example

    y=x2−2x+5:  a=1>0,  b2−4ac=−16<0y = x^2 - 2x + 5:\; a = 1 > 0,\; b^2 - 4ac = -16 < 0

    A concrete case to test the reasoning against.

  13. Complete the square for the example

    x2−2x+5=(x−1)2+4,q=4>0x^2 - 2x + 5 = (x - 1)^2 + 4,\quad q = 4 > 0

    As predicted: qq is positive and the minimum is above the axis.

  14. See it on the graph

    minimum (1, 4) lies above the x-axis\text{minimum } (1,\ 4) \text{ lies above the } x\text{-axis}

    The example curve is U-shaped and sits entirely above the xx-axis.

  15. State the conclusion

    U-shaped, no real roots, entirely above the x-axis, with q>0\text{U-shaped, no real roots, entirely above the } x\text{-axis, with } q > 0

    Both facts — the shape and the sign of qq — follow from a>0a > 0 and a negative discriminant.

Answer
U-shaped, entirely above the x-axis; completed square a(x+p)2+q with q>0\text{U-shaped, entirely above the } x\text{-axis; completed square } a(x + p)^2 + q \text{ with } q > 0
Question 2
5 markschallenging
The equation x2+kx+(k+3)=0x^2 + kx + (k + 3) = 0 has a repeated root. Work out the possible values of kk.
Show worked solution

Worked solution

  1. Translate the condition

    repeated root  ⟺  b2−4ac=0\text{repeated root} \iff b^2 - 4ac = 0

    A repeated root means the discriminant is exactly zero.

  2. Write down aa, bb and cc

    a=1,b=k,c=k+3a = 1,\quad b = k,\quad c = k + 3

    The unknown kk appears in both bb and cc — that is what makes this harder.

  3. Form the equation

    k2−4(1)(k+3)=0k^2 - 4(1)(k + 3) = 0

    Substitute into b2−4ac=0b^2 - 4ac = 0, keeping (k+3)(k + 3) bracketed.

  4. Expand the bracket

    k2−4k−12=0k^2 - 4k - 12 = 0

    −4(k+3)=−4k−12-4(k + 3) = -4k - 12 — the −4-4 multiplies both terms.

  5. Recognise a quadratic in kk

    k2−4k−12=0k^2 - 4k - 12 = 0

    The condition itself is a quadratic equation, this time in kk.

  6. Look for the factor pair

    −6×2=−12,−6+2=−4-6 \times 2 = -12,\quad -6 + 2 = -4

    A pair multiplying to −12-12 and adding to −4-4.

  7. Factorise

    (k−6)(k+2)=0(k - 6)(k + 2) = 0

    Check by expanding: k2+2k−6k−12=k2−4k−12k^2 + 2k - 6k - 12 = k^2 - 4k - 12.

  8. Solve for kk

    k=6 or k=−2k = 6 \text{ or } k = -2

    Each bracket is set equal to zero in turn.

  9. Check k=6k = 6

    x2+6x+9=0x^2 + 6x + 9 = 0

    With k=6k = 6, c=6+3=9c = 6 + 3 = 9.

  10. Confirm the repeated root for k=6k = 6

    x2+6x+9=(x+3)2⇒x=−3x^2 + 6x + 9 = (x + 3)^2 \Rightarrow x = -3

    A perfect square, so the root is repeated — the condition holds.

  11. Check k=−2k = -2

    x2−2x+1=0x^2 - 2x + 1 = 0

    With k=−2k = -2, c=−2+3=1c = -2 + 3 = 1.

  12. Confirm the repeated root for k=−2k = -2

    x2−2x+1=(x−1)2⇒x=1x^2 - 2x + 1 = (x - 1)^2 \Rightarrow x = 1

    Also a perfect square, repeated root x=1x = 1 — the condition holds here too.

  13. Check both discriminants

    62−4(9)=0,(−2)2−4(1)=06^2 - 4(9) = 0,\qquad (-2)^2 - 4(1) = 0

    Both are zero, as required.

  14. See it on the graph

    the curve touches the x-axis at (−3, 0)\text{the curve touches the } x\text{-axis at } (-3,\ 0)

    With k=6k = 6 the parabola y=x2+6x+9y = x^2 + 6x + 9 touches the axis at a single point.

  15. State the answer

    k=6 or k=−2k = 6 \text{ or } k = -2

    Both values make the discriminant zero, giving a repeated root.

Answer
k=6 or k=−2k = 6 \text{ or } k = -2
Question 3
6 markschallenging
Write 2x2+5x+12x^2 + 5x + 1 in the form a(x+p)2+qa(x + p)^2 + q and hence solve 2x2+5x+1=02x^2 + 5x + 1 = 0, giving your answers in exact surd form.
Show worked solution

Worked solution

  1. Factor the coefficient of x2x^2 out of the first two terms

    2x2+5x+1=2(x2+52x)+12x^2 + 5x + 1 = 2\left(x^2 + \frac{5}{2}x\right) + 1

    The 22 must come out of the xx term as well as the x2x^2 term — 5÷2=525 \div 2 = \frac{5}{2}. Taking it out of the x2x^2 term only is the classic error, and it wrecks the constant later.

  2. Complete the square

    2x2+5x+1=2(x+54)2−1782x^2 + 5x + 1 = 2(x + \frac{5}{4})^2 - \frac{17}{8}

    Complete the square inside the bracket, then multiply the correction term back out by 22 and combine it with the constant.

  3. Rewrite the equation

    2(x+54)2−178=02(x + \frac{5}{4})^2 - \frac{17}{8} = 0

    The equation is unchanged — only the way the quadratic is written has changed.

  4. Isolate the square term

    2(x+54)2=1782(x + \frac{5}{4})^2 = \frac{17}{8}

    Move the constant across to the right-hand side.

  5. Divide by the coefficient

    (x+54)2=1716(x + \frac{5}{4})^2 = \frac{17}{16}

    Dividing both sides by 22 leaves the square on its own.

  6. Square root both sides

    x+54=±174x + \frac{5}{4} = \pm \frac{\sqrt{17}}{4}

    Remember the ±\pm: a positive and a negative square root both work, and that is exactly what gives the two solutions. Rationalising the denominator of the square root turns 1716\sqrt{\frac{17}{16}} into 174\frac{\sqrt{17}}{4}.

  7. Solve for xx

    x=−5±174x = \frac{-5 \pm \sqrt{17}}{4}

    Subtracting 54\frac{5}{4} from both sides gives the exact solutions.

  8. Check the discriminant agrees

    b2−4ac=(5)2−4(2)(1)=17b^2 - 4ac = (5)^2 - 4(2)(1) = 17

    17>017 > 0 and is not a square number, so two distinct irrational (surd) solutions are exactly what we should expect.

  9. Check the sum of the roots

    (−5−174)+(−5+174)=−52\left(\frac{-5 - \sqrt{17}}{4}\right) + \left(\frac{-5 + \sqrt{17}}{4}\right) = - \frac{5}{2}

    The two roots must add to −ba=−52-\frac{b}{a} = - \frac{5}{2}, and they do.

  10. Check the product of the roots

    (−5−174)(−5+174)=12\left(\frac{-5 - \sqrt{17}}{4}\right)\left(\frac{-5 + \sqrt{17}}{4}\right) = \frac{1}{2}

    The roots must multiply to ca=12\frac{c}{a} = \frac{1}{2}, and they do — a quick, reliable check on surd answers.

  11. Check by expanding the completed square

    2(x+54)2−178=2x2+5x+12(x + \frac{5}{4})^2 - \frac{17}{8} = 2x^2 + 5x + 1

    Expanding the completed square returns the original quadratic, so nothing was lost in the rewrite. In particular the 22 was correctly multiplied through the correction term.

  12. Why factorising would not have worked

    17 is not a perfect square17 \text{ is not a perfect square}

    Because the discriminant is not a square number there is no factorisation over the integers — completing the square (or the formula) is the only route to these roots.

  13. Read off the turning point

    (−54, −178)(- \frac{5}{4},\ - \frac{17}{8})

    The completed square hands over the turning point for free: the bracket is zero at x=−54x = - \frac{5}{4}, giving y=−178y = - \frac{17}{8}. It sits midway between the two roots.

  14. Decimal check

    x≈−2.28 or x≈−0.22x \approx -2.28 \text{ or } x \approx -0.22

    A decimal check confirms the surds are about the right size, but the exact surd form is what is asked for here.

  15. State the exact solutions

    x=−5±174x = \frac{-5 \pm \sqrt{17}}{4}

    These are the exact answers; no rounding has been used anywhere.

Answer
x=−5±174x = \frac{-5 \pm \sqrt{17}}{4}
Question 4
5 markschallenging
Solve 2x2−7x+2=02x^2 - 7x + 2 = 0. Give your answers correct to 22 decimal places.
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Worked solution

  1. Write down aa, bb and cc

    a=2,b=−7,c=2a = 2,\quad b = -7,\quad c = 2

    Read the coefficients straight off the equation, keeping the signs with the numbers.

  2. State the quadratic formula

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    This formula solves every quadratic, whether or not it factorises.

  3. Substitute the values

    x=−(−7)±(−7)2−4(2)(2)2(2)x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(2)}}{2(2)}

    Bracket every substituted value — that is what keeps the signs of a negative bb or cc under control.

  4. Work out the discriminant

    b2−4ac=49−(16)=33b^2 - 4ac = 49 - (16) = 33

    33>033 > 0, so there are two different real solutions.

  5. Evaluate the square root

    33=5.7445626…\sqrt{33} = 5.7445626\ldots

    Keep plenty of figures at this stage — rounding early is the classic way to lose the last decimal place.

  6. Work out the two solutions

    x=7+5.7445626…4=3.1861407…orx=7−5.7445626…4=0.31385934…x = \frac{7 + 5.7445626\ldots}{4} = 3.1861407\ldots \quad\text{or}\quad x = \frac{7 - 5.7445626\ldots}{4} = 0.31385934\ldots

    Take the ++ root and the −- root separately.

  7. Round as asked

    x=0.31 or x=3.19x = 0.31 \text{ or } x = 3.19

    Rounded to 2 decimal places.

  8. Keep the exact form as a check

    x=7±334x = \frac{7 \pm \sqrt{33}}{4}

    The exact surd form is worth writing down: it makes the rounding easy to check, and it is the answer if "exact" is ever asked for.

  9. Check the rounding of each root

    0.31385934…→0.31,3.1861407…→3.190.31385934\ldots \to 0.31,\quad 3.1861407\ldots \to 3.19

    Look at the digit after the last one being kept to decide whether to round up or down.

  10. Check the sum of the roots

    sum=72=−ba\text{sum} = \frac{7}{2} = -\frac{b}{a}

    The formula guarantees the roots add to −ba=72-\frac{b}{a} = \frac{7}{2} — a fast check that costs nothing.

  11. Check the product of the roots

    product=1=ca\text{product} = 1 = \frac{c}{a}

    The roots must multiply to ca=1\frac{c}{a} = 1, and they do.

  12. Substitute the larger root back

    2x2−7x+2=0 when x=7+3342x^2 - 7x + 2 = 0 \text{ when } x = \frac{7 + \sqrt{33}}{4}

    Substituting the exact root into the original quadratic gives zero, so it is genuinely a solution.

  13. Why not factorise?

    b2−4ac=33b^2 - 4ac = 33

    The discriminant is not a perfect square, so no integer factorisation exists — the formula (or completing the square) is needed.

  14. Complete the square as a cross-check

    2x2−7x+2=2(x−74)2−3382x^2 - 7x + 2 = 2(x - \frac{7}{4})^2 - \frac{33}{8}

    Completing the square and then square-rooting gives exactly the same two solutions — the quadratic formula is just this process carried out once, in general.

  15. State both answers

    x=0.31 or x=3.19x = 0.31 \text{ or } x = 3.19

    Both solutions must be given — a quadratic has two.

Answer
x=0.31 or x=3.19x = 0.31 \text{ or } x = 3.19
Question 5
6 markschallenging
The equation x2+6x+c=0x^2 + 6x + c = 0 has two real roots whose difference is 44. Work out the value of cc.
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Worked solution

  1. Complete the square

    x2+6x+c=(x+3)2−9+cx^2 + 6x + c = (x + 3)^2 - 9 + c

    Halve the 66 to get 33, subtract 99, then carry the cc down.

  2. Set it equal to zero

    (x+3)2−9+c=0(x + 3)^2 - 9 + c = 0

    The completed square makes the roots easy to describe.

  3. Isolate the square

    (x+3)2=9−c(x + 3)^2 = 9 - c

    Move the constants to the right-hand side.

  4. Square root both sides

    x+3=±9−cx + 3 = \pm\sqrt{9 - c}

    This needs 9−c≥09 - c \ge 0 for real roots.

  5. Write the two roots

    x=−3±9−cx = -3 \pm \sqrt{9 - c}

    The roots sit symmetrically either side of x=−3x = -3, the line of symmetry.

  6. Find the difference of the roots

    (−3+9−c)−(−3−9−c)=29−c\left(-3 + \sqrt{9 - c}\right) - \left(-3 - \sqrt{9 - c}\right) = 2\sqrt{9 - c}

    The −3-3 cancels; the gap between the roots is twice the surd.

  7. Use the given condition

    29−c=42\sqrt{9 - c} = 4

    The difference between the roots is 44.

  8. Divide by 2

    9−c=2\sqrt{9 - c} = 2

    Halve both sides.

  9. Square both sides

    9−c=49 - c = 4

    Squaring removes the surd.

  10. Solve for cc

    c=5c = 5

    Rearranging gives c=9−4=5c = 9 - 4 = 5.

  11. Check the roots

    x2+6x+5=0⇒(x+5)(x+1)=0x^2 + 6x + 5 = 0 \Rightarrow (x + 5)(x + 1) = 0

    With c=5c = 5 the equation factorises.

  12. Check the difference

    x=−1 or x=−5;−1−(−5)=4x = -1 \text{ or } x = -5;\quad -1 - (-5) = 4

    The roots differ by exactly 44, as required.

  13. Check the discriminant

    b2−4ac=36−20=16>0b^2 - 4ac = 36 - 20 = 16 > 0

    Positive, so there really are two distinct real roots — the condition is consistent.

  14. Note the general result

    difference of roots=b2−4aca\text{difference of roots} = \frac{\sqrt{b^2 - 4ac}}{a}

    Here 161=4\frac{\sqrt{16}}{1} = 4 — the difference of the roots is always the square root of the discriminant divided by aa.

  15. State the answer

    c=5c = 5

    The value of cc is 55.

Answer
c=5c = 5

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