Question

How do you solve ${x^2} + 3x = 0?$

Answer 1

Hint:First take out the common terms in the expression then you will get two factors, individually equate them with zero, you will get the solution. As the given equation is of two degrees you should get two solutions for $x$.

Complete step by step answer:
In order to solve the given equation ${x^2} + 3x = 0$ we will simply take out the common terms first from the equation as follows:
${x^2} + 3x = 0 \\
\Rightarrow x(x + 3) = 0 $
Here we have taken out $x$ from both the terms, and as the result we get two factors, now equating both of them with zero individually in order to get the required solution,
$x = 0\;{\text{and}}\;(x + 3) = 0 \\
\Rightarrow x = 0\;{\text{and}}\;x = - 3 \\ $
Therefore $x = 0{\kern 1pt} \;{\text{and}}\;x = - 3$ are the required solutions of the expression ${x^2} + 3x = 0$.


Note:When solving the equation take care of the point that you have not cancelled the variable which you have taken common, which will result in only one solution that is $x = - 3$ and the solution $x = 0$ will vanish. This question can be solved by one more way that is with the help of quadratic formula as follows:
To solve the expression ${x^2} + 3x = 0$ with help of quadratic formula, first find the values of respective quadratic coefficients of the given expressionTo find quadratic coefficients of the expression comparing it with the standard quadratic equation
${x^2} + 3x = 0\;{\text{and}}\;a{x^2} + bx + c$
We can also write it as
${x^2} + 3x + 0 = 0\;{\text{and}}\;a{x^2} + bx + c$
Therefore respective values of quadratic coefficients are
$a = 1,\;b = 3\;{\text{and}}\;c = 0$
Now we know that the solution for the quadratic equation $a{x^2} + bx + c = 0$ is given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D$ is the discriminant of the quadratic expression, which can be calculated as
$D = \sqrt {{b^2} - 4ac} $
So first finding the value of discriminant,
$D = {3^2} - 4 \times 1 \times 0 \\
\Rightarrow D = 9 \\ $
Now substituting all the respective values in quadratic formula in order to get the solution for $x$
$x = \dfrac{{ - 3 \pm \sqrt 9 }}{{2 \times 1}} \\
\Rightarrow x = \dfrac{{ - 3 \pm 3}}{2} \\ $
Therefore $x = \dfrac{{ - 3 + 3}}{2} = \dfrac{0}{2} = 0\;{\text{and}}\;x = \dfrac{{ - 3 - 3}}{2} = \dfrac{{ - 6}}{2} = - 3$ are the required solutions for the equation ${x^2} + 3x = 0$