Completing square and formula Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Completing square and formula questions. See exactly how to solve problems on solving a completed square, square rooting both sides, surd form, completing the square.

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

Worked solution

Isolate the square
$(x + 3)^2 = 25$
The squared bracket is already on its own, so the square root can be taken next.
Square root both sides
$x + 3 = \pm 5$
Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.
State the solutions
$x = -8 \text{ or } x = 2$
Subtract the number inside the bracket from both sides to finish.

Answer

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

Question 2

1 markeasy

Solve

(x - 1)^2 = 16

Worked solution

Isolate the square
$(x - 1)^2 = 16$
The squared bracket is already on its own, so the square root can be taken next.
Square root both sides
$x - 1 = \pm 4$
Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.
State the solutions
$x = -3 \text{ or } x = 5$
Subtract the number inside the bracket from both sides to finish.

Answer

x = -3 \text{ or } x = 5

Question 3

2 markseasy

Solve

(x - 2)^2 = 7

. Give your answers in exact surd form.

Worked solution

Isolate the square
$(x - 2)^2 = 7$
The squared bracket is already on its own, so the square root can be taken next.
Square root both sides
$x - 2 = \pm \sqrt{7}$
Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.
State the solutions
$x = 2 \pm \sqrt{7}$
Subtract the number inside the bracket from both sides to finish.

Answer

x = 2 \pm \sqrt{7}

Question 4

2 markseasy

Solve

(x + 5)^2 - 3 = 0

. Give your answers in exact surd form.

Worked solution

Isolate the square
$(x + 5)^2 = 3$
Add 3 to both sides so the squared bracket is on its own.
Square root both sides
$x + 5 = \pm \sqrt{3}$
Both a positive and a negative square root satisfy the equation — this is where the two solutions come from.
State the solutions
$x = -5 \pm \sqrt{3}$
Subtract the number inside the bracket from both sides to finish.

Answer

x = -5 \pm \sqrt{3}

Question 5

1 markeasy

Write

x^2 + 4x

in the form

(x + p)^2 + q

Worked solution

Halve the coefficient of $x$
$\frac{4}{2} = 2$
The number inside the bracket is always half the coefficient of $x$ , so $p = 2$ .
Expand to see the extra term
$(x + 2)^2 = x^2 + 4x + 4$
The square carries a spare $+4$ that was not in the original expression, so $4$ has to be subtracted again.
State the completed square
$(x + 2)^2 - 4$
So $p = 2$ and $q = -4$ .

Answer

(x + 2)^2 - 4

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