Question

How do you solve $x^2-4x-9=0$ by completing the square?

Answer 1

Hint: In this particular problem, we have to solve the given quadratic equation which is in the form of $a{x}^2+bx+c=0$ using the method of completing the square. In this method, usually we add the square of half of the x-term on both sides of the equation to find the roots of the equation. Roots are those values which satisfy the equation.

Complete step-by-step solution:
Now, let’s begin to solve the question.
As we already know that quadratic equation is of the form $a{x}^2+bx+c-0$ with the highest degree as 2. It has two roots. Roots are those values which satisfy the equation. There are various methods of finding the roots of quadratic equations. Those methods are: middle term splitting, by using quadratic formula or by completing the square method. In this question, it is asked to solve by completing the square method. In this method, first we multiply the coefficient of x-term with half and then square it and add that result on both the sides of the equation. Let’s see how it can be done.
First write the given equation.
$\Rightarrow x^2-4x-9=0$
Now, add ${({\displaystyle \frac{1}{2}}coefficientofx-term)}^2$ on both the sides.
So,
 $\Rightarrow {({\displaystyle \frac{1}{2}}coefficientofx-term)}^2$= ${({\displaystyle \frac{1}{2}}\times (-4))}^2={(-2)}^2=4$
Add 4 on both the sides of the equation:
$\Rightarrow x^2-4x+4-9=0+4....(i)$
Now, we can see that $x^2-4x+4$ can be factored by middle term splitting:
$\begin{array}{rl}& \Rightarrow x^2-2x-2x+4\\ & \Rightarrow x(x-2)-2(x-2)\\ & \Rightarrow (x-2)(x-2)\iff {(x-2)}^2\end{array}$
Place the factored value in the equation(i), we will get:
$\Rightarrow {(x-2)}^2-9=4$
Solve the constant terms:
$\begin{array}{rl}& \Rightarrow {(x-2)}^2=4+9\\ & \Rightarrow {(x-2)}^2=13\end{array}$
Now, remove the square, it will become the under root for 13:
$\Rightarrow x-2=\pm \sqrt{13}$
So roots of x will be:
$\Rightarrow x=2+\sqrt{13},2+\sqrt{13}$
This is the answer.

Note: Always remember that on applying the square method, we have to perform middle term splitting after adding the square of the coefficient of x-term. We have placed the $\pm$ sign before $\sqrt{13}$ because we have two roots which is a positive and a negative one. So the sign is required.