Question

find the positive root of square root 3 X square + 6 = 9
$find \: the \: positive \: root \: of \sqrt{3x {}^{2} } + 6 = 9$
​

Answer 1

Ans : $x=\sqrt{3}$

$\sqrt{3x^2}+6=9$

Square both sides.

$(\sqrt{3x^2})^2=3^2$

$3x^2=9$

$x^2=3$

Root check!

$\sqrt{3\times 3}+6=9$

Ok, two roots are both true.

The roots of $x^2=3$ are $x=\pm\sqrt{3}$. (This is the definition of square root.)

The positive root is $x=\sqrt{3}$.

Answer 2

$\sqrt{3 {x}^{2} } + 6 = 9 \\ \sqrt{3 {x}^{2} } = 3 \\ squaring \: both \: sides \\ \\ 3 {x}^{2} = 9 \\ {x}^{2} = 3 \\ x = + \sqrt{3}$

Here the value of x can be both +√3 and -√3 but since we take the positive answer the answer to this question is +√3

Hope this is helpful to you

Pls mark as brainliest

Follow me and I'll follow back