Question

How do you solve the equation: $x = \sqrt {3x + 40}$?

Answer 1

First, convert the given equation into a standard quadratic equation. For this, we have to remove the radical on the right side of the equation, by squaring both sides of the equation. Next, move all terms to the left side of the equation by subtracting them 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 solution:
Given equation: $x = \sqrt {3x + 40}$
We have to convert the given equation into a standard quadratic equation.
For this, we have to remove the radical on the right side of the equation, by squaring both sides of the equation.
${x^2} = 3x + 40$
Now, move $3x + 40$ to the left side of the equation by subtracting $3x + 40$ from both sides of the equation.
${x^2} - 3x - 40 = 0$
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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Next, compare ${x^2} - 3x - 40 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - 3x - 40 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 3$ and $c = - 40$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 3} \right)^2} - 4\left( 1 \right)\left( { - 40} \right)$
After simplifying the result, we get
$\Rightarrow D = 9 + 160$
$\Rightarrow D = 169$
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
$x = \dfrac{{ - \left( { - 3} \right) \pm 13}}{{2 \times 1}}$
It can be written as
$\Rightarrow x = \dfrac{{3 \pm 13}}{2}$
$\Rightarrow x = 8$ and $x = - 5$
So, $x = 8$ and $x = - 5$ are roots/solutions of equation ${x^2} - 3x - 40 = 0$.
Now, exclude the solutions that do not make $x = \sqrt {3x + 40}$ true.
$\therefore x = 8$
Therefore, $x = 8$ is the solution of the given equation.

Note: We can check our answer by putting $x = 8$ in the given equation.
$LHS = 8$
$RHS = \sqrt {24 + 40} = \sqrt {64} = 8$
$\therefore LHS = RHS$
Students often make mistakes in this type of problem by including all solutions of quadratic equations. So, be careful to exclude the solutions which do not satisfy the given equation.