Question

How do you solve $5{x^2} = 240?$

Answer 1

Hint:First divide both sides (left hand side and right hand side) with coefficient of ${x^2}$ then take the constant in the side where ${x^2}$ lies (i.e. normally left hand side) with the use of algebraic operation. And express the constant as square of a number then use the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ in order to factorize the expression and then individually compare the factors to zero to get the solution.

Complete step by step answer:
To solve the given equation $5{x^2} = 240$, we will first divide both sides with the coefficient of ${x^2}$ that is equals to $5$ in the given equation
$
5{x^2} = 240 \\
\Rightarrow \dfrac{{5{x^2}}}{5} = \dfrac{{240}}{5} \\
\Rightarrow {x^2} = 48 \\ $
Now subtracting $48$ from both sides in order to take the constant to the left hand side,
$
\Rightarrow {x^2} = 48 \\
\Rightarrow {x^2} - 48 = 48 - 48 \\
\Rightarrow {x^2} - 48 = 0 \\ $
Now we will express $48$ as square of some number, so we have to find its square root
$\sqrt {48} = 4\sqrt 3 $
Therefore the above equation will be further written as follows
$
\Rightarrow {x^2} - 48 = 0 \\
\Rightarrow {x^2} - {\left( {4\sqrt 3 } \right)^2} = 0 \\ $
Using the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ to factorize the expression at left hand side,
$
\Rightarrow {x^2} - {\left( {4\sqrt 3 } \right)^2} = 0 \\
\Rightarrow \left( {x + 4\sqrt 3 } \right)\left( {x - 4\sqrt 3 } \right) = 0 \\ $
So here two cases are been formed, that is either the first factor equals zero or the second, so we will take both for the solution,
$
\Rightarrow \left( {x + 4\sqrt 3 } \right) = 0\;{\text{and}}\;\left( {x - 4\sqrt 3 } \right) = 0 \\
\Rightarrow x = - 4\sqrt 3 \;{\text{and}}\;x = 4\sqrt 3 \\
 $
Therefore $x = - 4\sqrt 3 \;{\text{and}}\;4\sqrt 3 $ are both the solution for the equation $5{x^2} = 240$, because both are satisfying the equation. You can check whether both are satisfying the equation or not by putting them in the equation.

Note:We get two solutions for the given equation because the equation is of two degrees and one variable. The degree of the equation is directly proportional or equal to the number of roots or solutions for that equation. Although sometimes you will get less number of solutions, this is because the rests are imaginary solutions of the equation.