Question

How do you find the roots for $f(x) = {x^2} - 4$?

Answer 1

Hint: TThis equation is the quadratic equation. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of ${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term.
To solve this equation, we will apply the quadratic formula for the quadratic equation.

Formula used:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
Here, $\sqrt {{b^2} - 4ac} $ is called the discriminant. And it is denoted by $\Delta $.
If $\Delta $ is greater than 0, then we will get two distinct and real roots.
If $\Delta $is less than 0, then we will not get real roots. In this case, we will get two complex numbers.
If $\Delta $ is equal to 0, then we will get two equal real roots.

Complete step-by-step answer:
Here, the quadratic equation is
$ \Rightarrow {x^2} - 4$
Let us compare the above expression with $a{x^2} + bx + c$.
Here, we get the value of ‘a’ is 1, the value of ‘b’ is 0, and the value of ‘c’ is -4.
Now, let us find the discriminant $\Delta $.
$ \Rightarrow \Delta = {b^2} - 4ac$
Let us substitute the values.
$ \Rightarrow \Delta = {\left( 0 \right)^2} - 4\left( 1 \right)\left( { - 4} \right)$
Simplify it.
$ \Rightarrow \Delta = 0 + 16$
Subtract the right-hand side.
$ \Rightarrow \Delta = 16$
Here, $\Delta $ is greater than 0, then we will get two real roots.
Now,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Put all the values.
$ \Rightarrow x = \dfrac{{ - \left( 0 \right) \pm \sqrt {16} }}{{2\left( 1 \right)}}$
That is equal to
$ \Rightarrow x = \dfrac{{0 \pm 4}}{2}$
Let us take out 2 as a common factor from the numerator.
$ \Rightarrow x = \dfrac{{ \pm 4}}{2}$
That is equal to,
$ \Rightarrow x = \pm 2$

Hence, the two factors are $2$ and$ - 2$.

Note:
We can also solve this question by another method.
First equate f(x) with 0.
$ \Rightarrow f\left( x \right) = 0$
Therefore,
$ \Rightarrow {x^2} - 4 = 0$
Let us add 4 on both sides.
$ \Rightarrow {x^2} - 4 + 4 = 0 + 4$
That is equal to,
$ \Rightarrow {x^2} = 4$
Now, let us apply the square root on both sides.
$ \Rightarrow \sqrt {{x^2}} = \sqrt 4 $
The square root of ${x^2}$ is x and the square root of 4 is$ \pm 2$.
$ \Rightarrow x = \pm 2$