Question

Find the positive root of $\sqrt {3{x^2} + 6} = 9$

Answer 1

Hint: The square root of the number “n” is the number when multiplied by itself and equals to “n”. For example square root of $\sqrt 9 = \sqrt {{3^2}} = 3$

Complete step by step solution:
Given that $\sqrt {3{x^2} + 6} = 9$
Squaring on both the sides of the equation –
${(\sqrt {3{x^2} + 6} )^2} = {9^2}$
As per the property – the square and the square root cancels each other.
$3{x^2} + 6 = 81$
Take all constants on the right hand side of the equation.
$3{x^2} = 81 - 6$ [Sign of the number changes when we change the side, positive changes to negative and negative changes to positive sign]
$3{x^2} = 75$
${x^2} = \dfrac{{75}}{3}$ [When number from multiplication changes side, it goes to the denominator from the numerator]
${x^2} = 25$
Take square root on both the sides –
$\sqrt {{x^2}} = \sqrt {25} $
As we know that – the square and the square root cancels each other.
$
  x = \pm 5 \\
  x = 5{\text{ or }}x = ( - 5) \\
 $
Here, $x = ( - 5)$ is not applicable.
As here, we are asked to find the positive square root, therefore $x = 5$ is the required answer.
Hence, the positive root of $\sqrt {3{x^2} + 6} = 9$ is $x = 5$

Note: The squares and the square roots are opposite to each other and so cancel each other. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2}$, generally it is denoted by n to the power two i.e. ${n^2}$. The perfect square is the number which can be expressed as the product of the two equal integers. For example: $9$, it can be expressed as the product of equal integers. $9 = 3 \times 3$