Question

How do you solve ${x^2} - 4 = 5$?

Answer 1

Hint: First, move the constant part from the right-hand side to the left-hand side. After that, use the method of completing the square to factorize 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:
Given equation is,
$ \Rightarrow {x^2} - 4 = 5$
Move 5 from right-hand side to left-hand side,
$ \Rightarrow {x^2} - 4 - 5 = 0$
Simplify the terms,
$ \Rightarrow {x^2} - 9 = 0$
We will factorize 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 ${x^2} - 9 = 0$,
$ \Rightarrow {x^2} - 2 \times x \times 0 - 9 = 0$
Which is in $2ax$ form.
Step 3: Add and subtract ${a^2}$.
Since \[a = 0\], we have
$ \Rightarrow {x^2} - 2 \times x \times 0 + {0^2} - {0^2} - 9 = 0$
Step 4: Use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$,
So, we have
$ \Rightarrow {\left( {x - 0} \right)^2} - 9 = 0$
Step 5: Write the constant term in ${b^2}$ the form
So, we have,
$ \Rightarrow {x^2} - {3^2} = 0$
Step 6: Use the identity ${\left( {a - b} \right)^2} = \left( {a - b} \right)\left( {a + b} \right)$
So, we have
$ \Rightarrow \left( {x - 3} \right)\left( {x + 3} \right) = 0$
Then by zero product property, we have
$ \Rightarrow x + 3 = 0$ or $x - 3 = 0$
Move constant part on the right side,
$ \Rightarrow x = - 3$ or $x = 3$

Hence, the value of $x$ is -3 and 3.

Note:
We can solve this question in alternate ways.
Given equation is,
$ \Rightarrow {x^2} - 4 = 5$
Move 5 from right-hand side to left-hand side,
$ \Rightarrow {x^2} - 4 - 5 = 0$
Simplify the terms,
$ \Rightarrow {x^2} - 9 = 0$
First, we will compare the given equation with the standard quadratic equation which is given by $a{x^2} + bx + c = 0$.
On comparing we get the values $a = 1,b = 0,c = - 9$.
Now, we know that the quadratic formula is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values in the above formula we get
$ \Rightarrow x = \dfrac{{0 \pm \sqrt {{0^2} - 4 \times 1 \times - 9} }}{{2 \times 1}}$
Now, on solving the obtained equation we get
$ \Rightarrow x = \dfrac{{ \pm \sqrt {36} }}{2}$
Now, we know that the value of the square root $\sqrt {36} = 6$.
$ \Rightarrow x = \pm \dfrac{6}{2}$
Cancel out the common factors,
$ \Rightarrow x = \pm 3$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$ \Rightarrow x = - 3$ and $x = 3$
Hence, the value of $x$ is -3 and 3.