GCSE Completing square and formula Practice Questions

Free GCSE Completing square and formula practice questions with full step-by-step worked solutions. Covers solving a completed square, square rooting both sides, surd form, completing the square. Practise exam-style problems and check your method.

solving a completed squaresquare rooting both sidessurd formcompleting the squaremonic quadraticquadratic formula
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Solve (x+3)2=25(x + 3)^2 = 25.
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Worked solution

  1. Isolate the square

    (x+3)2=25(x + 3)^2 = 25

    The squared bracket is already on its own, so the square root can be taken next.

  2. Square root both sides

    x+3=±5x + 3 = \pm 5

    Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.

  3. State the solutions

    x=−8 or x=2x = -8 \text{ or } x = 2

    Subtract the number inside the bracket from both sides to finish.

Answer
x=−8 or x=2x = -8 \text{ or } x = 2
Question 2
2 markseasy
Solve (x+1)2=12(x + 1)^2 = 12. Give your answers in exact surd form, simplified as far as possible.
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Worked solution

  1. Isolate the square

    (x+1)2=12(x + 1)^2 = 12

    The squared bracket is already on its own, so the square root can be taken next.

  2. Square root both sides

    x+1=±23x + 1 = \pm 2\sqrt{3}

    Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.

  3. State the solutions

    x=−1±23x = -1 \pm 2\sqrt{3}

    Subtract the number inside the bracket from both sides to finish.

Answer
x=−1±23x = -1 \pm 2\sqrt{3}
Question 3
2 marksintermediate
Solve x2+2x−11=0x^2 + 2x - 11 = 0 by completing the square. Give your answers in exact surd form, simplified as far as possible.
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Worked solution

  1. Complete the square

    x2+2x−11=(x+1)2−12x^2 + 2x - 11 = (x + 1)^2 - 12

    Halve the coefficient of xx and subtract the square of it.

  2. Rewrite the equation

    (x+1)2−12=0(x + 1)^2 - 12 = 0

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

  3. Isolate the square

    (x+1)2=12(x + 1)^2 = 12

    Add 1212 to both sides.

  4. Square root both sides

    x+1=±23x + 1 = \pm 2\sqrt{3}

    Remember the ±\pm: a positive and a negative square root both work, and that is exactly what gives the two solutions. 12\sqrt{12} simplifies to 232\sqrt{3}.

  5. Solve for xx

    x=−1±23x = -1 \pm 2\sqrt{3}

    Subtracting 11 from both sides gives the exact solutions.

  6. State the exact solutions

    x=−1±23x = -1 \pm 2\sqrt{3}

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

Answer
x=−1±23x = -1 \pm 2\sqrt{3}
Question 4
4 markshard
The equation 2x2+6x+k=02x^2 + 6x + k = 0 has two distinct real solutions. Find the range of possible values of kk.
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Worked solution

  1. Translate the condition

    two distinct real solutions  ⟺  b2−4ac>0\text{two distinct real solutions} \iff b^2 - 4ac > 0

    Two *different* real roots means a strictly positive discriminant.

  2. Write down aa, bb and cc

    a=2,b=6,c=ka = 2,\quad b = 6,\quad c = k

    This time the unknown is the constant term.

  3. Form the inequality

    62−4(2)(k)>06^2 - 4(2)(k) > 0

    Substitute into b2−4ac>0b^2 - 4ac > 0.

  4. Simplify

    36−8k>036 - 8k > 0

    4×2=84 \times 2 = 8, so 4ac=8k4ac = 8k.

  5. Rearrange

    36>8k36 > 8k

    Add 8k8k to both sides — keeping kk positive avoids having to flip the inequality later.

  6. Divide by 8

    368>k\frac{36}{8} > k

    Dividing by a positive number leaves the inequality sign the same way round.

  7. Simplify the fraction

    k<92k < \frac{9}{2}

    368=92=4.5\frac{36}{8} = \frac{9}{2} = 4.5.

  8. Test a value inside the range

    k=0:  2x2+6x=0⇒b2−4ac=36>0k = 0:\; 2x^2 + 6x = 0 \Rightarrow b^2 - 4ac = 36 > 0

    Two distinct roots (x=0x = 0 and x=−3x = -3) — the condition holds.

  9. Test the boundary

    k=92:  b2−4ac=36−36=0k = \frac{9}{2}:\; b^2 - 4ac = 36 - 36 = 0

    At k=4.5k = 4.5 the roots become equal, not distinct, so the boundary is excluded — the inequality is strict.

  10. State the answer

    k<92k < \frac{9}{2}

    Any kk below 4.54.5 gives two distinct real solutions.

Answer
k<92k < \frac{9}{2}
Question 5
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

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