\({x}^{2}+10x2=0\)
2.2 Completing the square
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2.3 Quadratic formula

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 \((xb)(x+b)\). This simple factorisation leads to another technique for solving quadratic equations known as completing the square.
Consider the equation \(x^22x1=0\). We cannot easily factorise this expression. When we expand the perfect square \((x1)^2\) and examine the terms we see that \((x1)^2 = x^22x+1\).
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^22x1 &= 0 \\ (x^22x+1)11 &= 0 \\ (x^22x+1)2 &= 0 \\ (x1)^22 &= 0 \end{align*}
Method 1: Take square roots on both sides of the equation to solve for \(x\). \begin{align*} (x1)^22 &= 0 \\ (x1)^2 &= 2 \\ \sqrt{(x1)^2} &= \pm \sqrt{2} \\ x1 &= \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*} (x1)^22 &= 0 \\ (x1)^2  \left( \sqrt{2} \right)^2 &= 0 \\ \left( (x1) + \sqrt{2} \right)\left( (x1)  \sqrt{2} \right) &= 0 \end{align*}
The solution is then \begin{align*} (x1) + \sqrt{2} &= 0 \\ x &= 1  \sqrt{2} \end{align*} or \begin{align*} (x1)  \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: \(x^210x11=0\)
The equation is already in the form \(ax^2 + bx + c = 0\)
Make sure the coefficient of the \(x^2\) term is equal to \(\text{1}\)
\[x^210x11=0\]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 \\ (x5)^2  36 &= 0 \end{align*}Method 1: Take square roots on both sides of the equation
\begin{align*} (x5)^2  36 &= 0 \\ (x5)^2 &= 36 \\ x5 &= \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*} (x5)^2  (6)^2 &= 0 \\ \left[\left(x5\right) + 6\right] \left[\left(x5\right)  6\right] &= 0 \end{align*}Simplify and solve for \(x\)
\begin{align*} (x+1)(x11) &= 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
Solve the following equations by completing the square:
\({x}^{2}+4x+3=0\)
\(p^2  5 =  8p\)
\(2(6x + x^2) = 4\)
\({x}^{2}+5x+9=0\)
\(t^2 + 30 = 2(108t)\)
\(3{x}^{2}+6x2=0\)
\({z}^{2}+8z6=0\)
\(2z^2 = 11z\)
\(5+4z{z}^{2}=0\)
Solve for \(k\) in terms of \(a\): \(k^2 + 6k+ a = 0\)
Solve for \(y\) in terms of \(p\), \(q\) and \(r\): \(py^2 + qy + r = 0\)
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