Question

Find the values of x satisfying the equation ${{x}^{2}}-9=0$

Answer 1

Hint: Use the method of completing the square to factorise the LHS of the equation. Use zero product property which states that if ab = 0 , then a =0 or b =0 and hence find the value of x satisfying the equation.

Complete step-by-step answer:

We have
${{x}^{2}}-9=0$
We will factorise the LHS using completing the square method.
Step 1: Make the coefficient of ${{x}^{2}}$ as 1
Here coefficient of ${{x}^{2}}$ is already 1, so we move to the next step:
Step 2: Write the term containing x in 2ax form
We have
$\begin{align}
  & {{x}^{2}}-9=0 \\
 & \Rightarrow {{x}^{2}}-2\times 0\times x-9=0 \\
\end{align}$
Which is in 2ax form.
Step 3: Add and subtract ${{a}^{2}}$.
Since a =0, we have
${{x}^{2}}-2\times 0\times x+{{0}^{2}}-{{0}^{2}}+9=0$
Step 4: Use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$
Hence, we have
${{\left( x-0 \right)}^{2}}-9=0$
Step 5: Write the constant term in ${{b}^{2}}$ form
We have
$\begin{align}
  & {{\left( x-0 \right)}^{2}}-9=0 \\
 & \Rightarrow {{x}^{2}}-{{3}^{2}}=0 \\
\end{align}$
Step 6: Use the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Hence, we have
$\left( x-3 \right)\left( x+3 \right)=0$
Hence by zero product property, we have
$x-3=0\text{ or }x+3=0$
Hence, we have x =3 or x =-3.

Note: Alternative Solution:
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a = 1, b =0 and c = -9
Hence, we have
$x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times \left( 1 \right)\times \left( -9 \right)}}{2\times 1}=\dfrac{\pm 6}{2}=\pm 3$, which is the same as obtained above
Hence the roots of the equation are x = 3 and x = -3.