2.2 Completing the square (EMBFJ)
Completing the square
Can you solve each equation using two different methods?
- \(x^2 - 4 = 0\)
- \(x^2 - 8 = 0\)
- \(x^2 -4x + 4 = 0\)
- \(x^2 -4x - 4 = 0\)
Factorising the last equation is quite difficult. Use the previous examples as a hint and try to create a
difference of two squares.
We have seen that expressions of the form \(x^2 - b^2\) are known as differences of squares and can be factorised
as \((x-b)(x+b)\). This simple factorisation leads to another technique for solving quadratic equations known as
completing the square.
Consider the equation \(x^2-2x-1=0\). We cannot easily factorise this expression. When we expand the perfect
square \((x-1)^2\) and examine the terms we see that \((x-1)^2 = x^2-2x+1\).
We compare the two equations and notice that only the constant terms are different. We can create a perfect
square by adding and subtracting the same amount to the original equation.
\begin{align*}
x^2-2x-1 &= 0 \\
(x^2-2x+1)-1-1 &= 0 \\
(x^2-2x+1)-2 &= 0 \\
(x-1)^2-2 &= 0
\end{align*}
Method 1: Take square roots on both sides of the equation to solve for \(x\).
\begin{align*}
(x-1)^2-2 &= 0 \\
(x-1)^2 &= 2 \\
\sqrt{(x-1)^2} &= \pm \sqrt{2} \\
x-1 &= \pm \sqrt{2} \\
x &= 1 \pm \sqrt{2} \\
\text{Therefore }x &= 1 + \sqrt{2} \text{ or }x = 1 - \sqrt{2}
\end{align*}
Very important: Always remember to include both a positive and a negative answer when taking the
square root, since \(2^2 = 4\) and \((-2)^2 = 4\).
Method 2: Factorise the expression as a difference of two squares using \(2 =
\left(\sqrt{2}\right)^2\).
We can write
\begin{align*}
(x-1)^2-2 &= 0 \\
(x-1)^2 - \left( \sqrt{2} \right)^2 &= 0 \\
\left( (x-1) + \sqrt{2} \right)\left( (x-1) - \sqrt{2} \right) &= 0
\end{align*}
The solution is then
\begin{align*}
(x-1) + \sqrt{2} &= 0 \\
x &= 1 - \sqrt{2}
\end{align*}
or
\begin{align*}
(x-1) - \sqrt{2} &= 0 \\
x &= 1 + \sqrt{2}
\end{align*}
Method for solving quadratic equations by completing the square
-
Write the equation in the standard form \(a{x}^{2}+bx+c=0\).
-
Make the coefficient of the \({x}^{2}\) term equal to \(\text{1}\) by dividing the entire equation by \(a\).
-
Take half the coefficient of the \(x\) term and square it; then add and subtract it from the equation so that
the equation remains mathematically correct. In the example above, we added \(\text{1}\) to complete the
square and then subtracted \(\text{1}\) so that the equation remained true.
-
Write the left hand side as a difference of two squares.
-
Factorise the equation in terms of a difference of squares and solve for \(x\).
Worked example 6: Solving quadratic equations by completing the square
Solve by completing the square: \(x^2-10x-11=0\)
The equation is already in the form \(ax^2 + bx + c = 0\)
Make sure the coefficient of the \(x^2\) term is equal to \(\text{1}\)
\[x^2-10x-11=0\]
Take half the coefficient of the \(x\) term and square it; then add and subtract it from the
equation
The coefficient of the \(x\) term is \(-\text{10}\). Half of the coefficient of the \(x\) term is
\(-\text{5}\) and the square of it is \(\text{25}\). Therefore \(x^2 - 10x + 25 - 25 - 11 = 0\).
Write the trinomial as a perfect square
\begin{align*}
(x^2 - 10x + 25) - 25 - 11 &= 0 \\
(x-5)^2 - 36 &= 0
\end{align*}
Method 1: Take square roots on both sides of the equation
\begin{align*}
(x-5)^2 - 36 &= 0 \\
(x-5)^2 &= 36 \\
x-5 &= \pm\sqrt{36}
\end{align*}
Important: When taking a square root always remember that there is a positive and negative
answer, since \((6)^2 = 36\) and \((-6)^2 = 36\).
\[x - 5 = \pm 6\]
Solve for \(x\)
\[x = -1 \text{ or } x = 11\]
Method 2: Factorise equation as a difference of two squares
\begin{align*}
(x-5)^2 - (6)^2 &= 0 \\
\left[\left(x-5\right) + 6\right] \left[\left(x-5\right) - 6\right] &= 0
\end{align*}
Simplify and solve for \(x\)
\begin{align*}
(x+1)(x-11) &= 0 \\
\therefore x = -1 \text{ or } x &= 11
\end{align*}
Write the final answer
\[x = -1 \text{ or } x = 11\]
Notice that both methods produce the same answer. These roots are rational because \(\text{36}\) is a perfect
square.
Worked example 7: Solving quadratic equations by completing the square
Solve by completing the square: \(2x^2 - 6x - 10 = 0\)
The equation is already in standard form \(a{x}^{2}+bx+c=0\)
Make sure that the coefficient of the \(x^2\) term is equal to \(\text{1}\)
The coefficient of the \({x}^{2}\) term is \(\text{2}\). Therefore divide the entire equation by
\(\text{2}\):
\[x^2 - 3x - 5 = 0\]
Take half the coefficient of the \(x\) term, square it; then add and subtract it from the
equation
The coefficient of the \(x\) term is \(-\text{3}\), so then \(\left( \dfrac{-3}{2} \right)^2 =
\dfrac{9}{4}\):
\[\left( x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} - 5 = 0\]
Write the trinomial as a perfect square
\begin{align*}
\left( x - \frac{3}{2} \right)^2 - \frac{9}{4} - \frac{20}{4} &= 0 \\
\left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0
\end{align*}
Method 1: Take square roots on both sides of the equation
\begin{align*}
\left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\
\left( x - \frac{3}{2} \right)^2 &= \frac{29}{4} \\
x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}}
\end{align*}
Remember: When taking a square root there is a positive and a negative answer.
Solve for \(x\)
\begin{align*}
x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}} \\
x &= \frac{3}{2} \pm \frac{\sqrt{29}}{2} \\
&= \frac{3 \pm \sqrt{29}}{2}
\end{align*}
Method 2: Factorise equation as a difference of two squares
\begin{align*}
\left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\
\left( x - \frac{3}{2} \right)^2 - \left( \sqrt{\frac{29}{4}} \right)^2 &= 0\\
\left( x - \frac{3}{2} - \sqrt{\frac{29}{4}} \right) \left( x - \frac{3}{2} + \sqrt{\frac{29}{4}} \right) &=
0
\end{align*}
Solve for \(x\)
\begin{align*}
\left( x - \frac{3}{2} - \frac{\sqrt{29}}{2} \right) \left( x - \frac{3}{2} + \frac{\sqrt{29}}{2} \right) &=
0 \\
\text{Therefore } x = \frac{3}{2} + \frac{\sqrt{29}}{2} &\text{ or } x = \frac{3}{2} - \frac{\sqrt{29}}{2}
\end{align*}
Notice that these roots are irrational since \(\text{29}\) is not a perfect square.
Solution by completing the square
Textbook Exercise 2.2
\begin{align*}
x^2 + 10x - 2 &= 0\\
x^2 +10x&= 2\\
x^2 +10x +25 &= 2 + 25\\
(x+5)^2 - 27 &= 0\\
\left[(x+5) + \sqrt{27}\right]\left[(x+5) - \sqrt{27}\right] &= 0\\
(x+5) = -\sqrt{27} &\text{ or } (x+5) = \sqrt{27}\\
x = -5 - 3\sqrt{3} &\text{ or } x= -5 + 3\sqrt{3}
\end{align*}
\begin{align*}
x^2+4x+3&=0\\
x^2 +4x &= -3\\
x^2 + 4x + 4 &= -3 + 4\\
(x+2)^2 &= 1\\
(x+2)&= \pm \sqrt {1}\\
&=\pm 1\\
x = -2 + 1 = -1 &\text{ or } x = -2-1=-3
\end{align*}
\begin{align*}
p{^2} - 5 &= - 8p\\
p{^2}+8p-5 &= 0\\
p^2 + 8p &=5\\
p^2 + 8p + 16 &= 5 + 16\\
(p+4)^2 &= 21\\
(p+4)&=\pm \sqrt{21}\\
p &= -4 \pm \sqrt{21}
\end{align*}
\begin{align*}
2(6x + x^2) &= -4 \\
2x^2+12x+4 &= 0\\
x^2 + 6x + 2 &= 0\\
x^2 +6x &= -2\\
x^2 + 6x + 9&= -2 +9\\
(x+3)^2 &= 7\\
x + 3 &= \pm \sqrt{7}\\
x &= -3 \pm \sqrt{7}
\end{align*}
\begin{align*}
x^2+5x+9 &= 0\\
x^2 + 5x &= -9\\
x^2 + 5x + \frac{25}{4} &= -9 + \frac{25}{4}\\
\left(x+\frac{5}{2}\right)^2 &= -\frac{11}{4}\\
x + \frac{5}{2} &= \pm \sqrt{-\frac{11}{4}}\\
\text{ No real solution }
\end{align*}
\begin{align*}
t^2 + 30 &= 2(10 - 8t) \\
t^2+16t+10&=0\\
t^2 +16t &= -10\\
t^2 + 16t+64 &= -10 +64\\
(t+8)^2&=54\\
t + 8 &= \pm \sqrt{54}\\
t &= -8 \pm \sqrt{9 \times 6} \\
\therefore t &= -8 \pm 3\sqrt{6}
\end{align*}
\begin{align*}
3x^2+6x-2&=0\\
x^2 + 2x &= \frac{2}{3}\\
x^2 + 2x + 1 &= \frac{2}{3} + 1\\
(x+1)^2 &= \frac{5}{3}\\
x +1 &= \pm \sqrt{\frac{5}{3}}\\
x &= -1 \pm \sqrt{\frac{5}{3}}
\end{align*}
\begin{align*}
z^2 + 8z - 6 &= 0\\
z^2 +8z &= 6\\
z^2 + 8z+16 &= 6+16\\
(z+4)^2 &= 22\\
z+4 &= \pm \sqrt{22}\\
z &= -4 \pm \sqrt{22}
\end{align*}
\begin{align*}
2z^2 &= 11z\\
2z^2 - 11z &= 0\\
z^2 - \frac{11}{2}z &= 0\\
z^2 - \frac{11}{2}z + \frac{121}{16} &= \frac{121}{16}\\
\left(z-\frac{11}{4}\right)^2 &= \frac{121}{16}\\
z - \frac{11}{4} &= \pm \frac{11}{4}\\
z = \frac{11}{4} + \frac{11}{4} = \frac{11}{2} &\text{ or } z = \frac{11}{4} - \frac{11}{4} = 0
\end{align*}
\begin{align*}
5 + 4z - z^2 &= 0\\
z^2 -4z &= 5\\
z^2 - 4z +4 &= 5+4\\
(z-2)^2 &= 9\\
z-2 &= \pm \sqrt{9}\\
z = 2+3 = 5 &\text{ or } z = 2-3 = -1
\end{align*}
Solve for \(k\) in terms of \(a\): \(k^2 + 6k+ a = 0\)
\begin{align*}
k^2 + 6k+ a &= 0\\
k^2 + 6k &= -a\\
k^2 + 6k + 9 &= 9 - a\\
(k + 3)^2 &= 9 - a\\
k + 3 &= \pm \sqrt{9 - a}\\
k & = -3 \pm \sqrt{9 - a}
\end{align*}
Solve for \(y\) in terms of \(p\), \(q\) and \(r\): \(py^2 + qy + r = 0\)
\begin{align*}
py^2 + qy + r &= 0\\
y^2 + \frac{q}{p}y + \frac{r}{p} &= 0\\
y^2 + \frac{q}{p}y &= -\frac{r}{p}\\
y^{2} + \frac{q}{p}y + \left(\frac{q}{2p}\right)^{2} &= \left(\frac{q}{2p}\right)^{2} - \frac{r}{p}\\
\left(y + \frac{q}{2p}\right)^2 &= \frac{q^{2}}{4p^{2}} - \frac{r}{p} \\
y + \frac{q}{2p} &= \pm \sqrt{\frac{q^{2} - 4pr}{4p^{2}}}\\
y &= -\frac{q}{2p} \pm \frac{\sqrt{q^{2} - 4pr}}{2p}\\
y & = \frac{-q \pm \sqrt{q^{2} - 4pr}}{2p}
\end{align*}
