Question

How do you find the roots for ${x^2} + 16x = 0?$

Answer 1

Hint: This is a two degree one variable equation, or simply a quadratic equation. We can find the roots with the help of a quadratic formula for that first find the discriminant of the given quadratic equation and then put the values directly in the roots formula of the quadratic equation to get the required solution. You will get two solutions for a quadratic equation.

Formula Used:For a quadratic equation $a{x^2} + bx + c = 0$, discriminant is given as follows
$D = {b^2} - 4ac$
And its roots are given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Use this information to solve the question.

Complete step by step solution:
In order to solve the given expression ${x^2} + 16x = 0$ with help of quadratic formula, we need to first simplify the equation as follows
\[
   \Rightarrow {x^2} + 16x = 0 \\
   \Rightarrow {x^2} + 16x + 0 = 0 \\
 \]
Now comparing it with standard quadratic equation to get value of quadratic coefficients
\[ \Rightarrow {x^2} + 16x + 0 = 0\;{\text{and}}\;a{x^2} + bx + c = 0\]
Therefore respective values of quadratic coefficients are
$a = 1,\;b = 16\;{\text{and}}\;c = 0$
Now we know that the solution for the quadratic equation $a{x^2} + bx + c = 0$ is given as
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D$ is the discriminant of the quadratic expression, which can be calculated as
$ \Rightarrow D = \sqrt {{b^2} - 4ac} $
So first finding the value of discriminant,
$
   \Rightarrow D = {16^2} - 4 \times 4 \times 0 \\
   \Rightarrow D = 256 \\
 $
Now substituting all the respective values in quadratic formula in order to get the solution for $x$
$
   \Rightarrow x = \dfrac{{ - 16 \pm \sqrt {256} }}{{2 \times 1}} \\
    \Rightarrow \dfrac{{ - 16 \pm 16}}{2} \\
    \Rightarrow \dfrac{{ - 16 + 16}}{2}\;{\text{or}}\;\dfrac{{ - 16 - 16}}{2} \\
   \Rightarrow 0\;{\text{or}}\; - 16 \\
 $

Therefore $x = 0\;{\text{and}}\;x = - 16$ are the required solutions for the equation ${x^2} + 16x = 0$


Note: Quadratic formula is only applicable for quadratic equations that are equations with degree two. Equations having degrees greater or less than two have other methods for finding roots. Sometimes students directly solve this type of question by cancelling out the degree of the variable by division and get the value, but in this process they left one solution for the equation that is $x = 0$, so never directly cancel the degree of variables.