Mathematics » Equations and Inequalities » Completing The Square

Completing the Square

 

Completing the Square

Optional Investigation

Completing the square

Can you solve each equation using two different methods?

  1. \(x^2 – 4 = 0\)
  2. \(x^2 – 8 = 0\)
  3. \(x^2 -4x + 4 = 0\)
  4. \(x^2 -4x – 4 = 0\)

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 \(2 = (\sqrt{2})^2\).

We can write

\begin{align*} (x-1)^2-2 &= 0 \\ (x-1)^2 – ( \sqrt{2} )^2 &= 0 \\ ( (x-1) + \sqrt{2} )( (x-1) – \sqrt{2} ) &= 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

  1. Write the equation in the standard form \(a{x}^{2}+bx+c=0\).

  2. Make the coefficient of the \({x}^{2}\) term equal to \(\text{1}\) by dividing the entire equation by \(a\).

  3. 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.

  4. Write the left hand side as a difference of two squares.

  5. Factorise the equation in terms of a difference of squares and solve for \(x\).

Example

Question

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 \\ [(x-5) + 6] [(x-5) – 6] &= 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.

Example

Question

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 \(( \dfrac{-3}{2} )^2 = \dfrac{9}{4}\):

\[( x^2 – 3x + \cfrac{9}{4}) – \cfrac{9}{4} – 5 = 0\]

Write the trinomial as a perfect square

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{9}{4} – \cfrac{20}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \end{align*}

Method 1: Take square roots on both sides of the equation

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 &= \cfrac{29}{4} \\ x – \cfrac{3}{2} &= \pm \sqrt{\cfrac{29}{4}} \end{align*}

Remember: When taking a square root there is a positive and a negative answer.

Solve for \(x\)

\begin{align*} x – \cfrac{3}{2} &= \pm \sqrt{\cfrac{29}{4}} \\ x &= \cfrac{3}{2} \pm \cfrac{\sqrt{29}}{2} \\ &= \cfrac{3 \pm \sqrt{29}}{2} \end{align*}

Method 2: Factorise equation as a difference of two squares

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 – ( \sqrt{\cfrac{29}{4}} )^2 &= 0\\ ( x – \cfrac{3}{2} – \sqrt{\cfrac{29}{4}} ) ( x – \cfrac{3}{2} + \sqrt{\cfrac{29}{4}} ) &= 0 \end{align*}

Solve for \(x\)

\begin{align*} ( x – \cfrac{3}{2} – \cfrac{\sqrt{29}}{2} ) ( x – \cfrac{3}{2} + \cfrac{\sqrt{29}}{2} ) &= 0 \\ \text{Therefore } x = \cfrac{3}{2} + \cfrac{\sqrt{29}}{2} &\text{ or } x = \cfrac{3}{2} – \cfrac{\sqrt{29}}{2} \end{align*}

Notice that these roots are irrational since \(\text{29}\) is not a perfect square.

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