Question

How do you solve $36{x^2} - 25 = 0$ ?

Answer 1

This is a quadratic equation, so it will have two solutions for $x$. Consider the coefficient of $x$ to be $0$, then the equation becomes $36{x^2} + 0x - 25 = 0$. Find the discriminant of the equation and solve it further as you solve a quadratic equation. Finally, you will get two solutions for $x$and both will be the required answer.

Complete answer:
Given equation $36{x^2} - 25 = 0$,
Let us consider the coefficient of $x$ to be $0$, then the equation will be written as
$36{x^2} + 0x - 25 = 0$_____ (I)
Now, we know that a general quadratic equation is written as
$a{x^2} + bx + c = 0$ .
And its solution can be written as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ _______ (II)
Where $D$ is the discriminant of the quadratic equation, and we can find the discriminant of a quadratic equation as shown below
$D = {b^2} - 4ac$ _______ (III)
Now, comparing the general quadratic equation to the given quadratic equation as written in equation (I), we get the values of $a,\;b$ and $c$ as
$a = 36,\;b = 0$ and $c = - 25$
Now putting these values in equation (III) to get the discriminant of the quadratic equation,
$\Rightarrow D = {b^2} - 4ac \\\ \Rightarrow D = {0^2} - 4 \times 36 \times ( - 25) \\\ \Rightarrow D = 0 + 3600 \\\ \Rightarrow D = 3600 \\\$
Now, putting the values of $a,\;b$ and $D$ in equation (II) to get the solution for $x$
$\Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\\ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt {3600} }}{{2 \times 36}} \\\ \Rightarrow x = \dfrac{{ \pm \sqrt {3600} }}{{2 \times 36}} \\\$
Here we know that $\sqrt {3600} = \pm 60$, so by solving further we will get,
$\Rightarrow x = \dfrac{{ \mp 60}}{{72}} \\\ \Rightarrow x = \dfrac{{ \mp 5}}{6} \\\$
We get the two solutions for $x$ as
$x = \dfrac{5}{6}$ and $x = \dfrac{{ - 5}}{6}$

Note: By finding just the discriminant of a quadratic equation, we can know about the nature of the roots or the characteristics of the solution as follows:
1. Roots will be real and distinct, if $D > 0$
2. Roots will be real and equal, if $D = 0$
3. Roots will be imaginary or no real roots, if $D < 0$
Where $D$ is the discriminant of the quadratic equation.