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# How do you solve the equation $\dfrac{2}{3}{x^2} - 4 = 12?$

Last updated date: 04th Mar 2024
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Hint: First of all take the given expression, take all the constants on one side and the term with variable on one side and then simplify the equation for the resultant required value for the unknown term “x”.

Complete step-by-step solution:
Take the given expression: $\dfrac{2}{3}{x^2} - 4 = 12$
Move constant on the right hand side of the equation. When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$\dfrac{2}{3}{x^2} = 12 + 4$
Simplify the expression on the right hand side of the equation.
$\dfrac{2}{3}{x^2} = 16$
The term in division on one side, if moved to the opposite side then it goes to the numerator part.
$2{x^2} = 16 \times 3$
Simplify the above equation-
$2{x^2} = 48$
The term multiplicative on one side if moved to the opposite side then it goes to the denominator part.
$\Rightarrow {x^2} = \dfrac{{48}}{2}$
Find factors of the term on the numerator.
$\Rightarrow {x^2} = \dfrac{{24 \times 2}}{2}$
Common factors from the numerator and the denominator cancels each other. Therefore, remove from the numerator and the denominator.
$\Rightarrow {x^2} = 24$
Take square root on both sides of the above equation.
$\Rightarrow \sqrt {{x^2}} = \sqrt {24}$
Square and square root cancel each other on the left hand side of the equation.
$\Rightarrow x = \sqrt {4 \times 6} \\ \Rightarrow x = \sqrt {{2^2} \times 6} \\ \Rightarrow x = \sqrt {{2^2}} \times \sqrt 6 \\$
Square and square root cancel in the above equation.
$\Rightarrow x = \pm 2\sqrt 6$
This is the required solution.

Note: Always remember that the square of positive or the negative number always gives us the positive number but the square root of positive number gives positive or the negative number therefore, we have kept plus or minus ahead of the number. Be careful about the plus or minus signs. Be good in multiples and remember at least twenty.