Question

How do you solve $4{x^2} = 144$?

Answer 1

Hint: First, move $144$ to the left side of the equation by subtracting $144$ from both sides of the equation. Next, compare the quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.
Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
First, move $144$ to the left side of the equation by subtracting $144$ from both sides of the equation.
$ \Rightarrow 4{x^2} - 144 = 0$
Now, compare $4{x^2} - 144 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $4{x^2} - 144 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 4$, $b = 0$ and $c = 144$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 0 \right)^2} - 4\left( 4 \right)\left( {144} \right)$
After simplifying the result, we get
$ \Rightarrow D = 2304$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{0 \pm 2 \times 2 \times 12}}{{2 \times 4}}$
Divide numerator and denominator by $8$, we get
$ \Rightarrow x = \pm 6$
$ \Rightarrow x = 6$ and $x = - 6$
So, $x = 6$ and $x = - 6$ are roots/solutions of equation $4{x^2} = 144$.
Final solution: Therefore, the solutions to the quadratic equation $4{x^2} = 144$ are $x = \pm 6$.

Note:
We can also find the solution of the quadratic equation $4{x^2} = 144$ by taking the square root of each side of the equation.
So, first divide both side of the equation by $4$, we get
$ \Rightarrow {x^2} = 36$
Now, take square root on both sides, we get
$ \Rightarrow x = \pm \sqrt {36} $
Simplify the right side of the equation.
Since, $\sqrt {36} $ can be written as $6$.
Thus, $x = \pm 6$
Therefore, $x = 6$ and $x = - 6$.
Final solution: Therefore, the solutions to the quadratic equation $4{x^2} = 144$ are $x = \pm 6$.