Question

Find the Value of x $\sqrt{ \bigg(5^0 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$​

Answer 1

$\large\bold{\underline{\underline{Given:-}}}$

$\sqrt{ \bigg(5^0 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

$\large\bold{\underline{\underline{To \: Find:-}}}$

The value of x

$\large\bold{\underline{\underline{Solution:-}}}$

$\sqrt{ \bigg(5^0 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

Rewriting 5⁰ = 1:

$\sqrt{ \bigg(1 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

$\sqrt{ \dfrac{5}{3} }= (0.6)^2 - 3x$

Rationalise the LHS by multiplying √3/√3:

$\dfrac{\sqrt{15} }{3} = (0.6)^2 - 3x$

Rewriting (0.6)² as a fraction:

$\dfrac{\sqrt{15} }{3} = \bigg( \dfrac{6}{10} \bigg)^2 - 3x$

$\dfrac{\sqrt{15} }{3} = \dfrac{9}{25} - 3x$

Multiply both sides by 3:

$\sqrt{15} = \dfrac{27}{25} - 9x$

Find x:

$9x = \dfrac{27}{25} - \sqrt{15}$

$9x = \dfrac{27 - 25\sqrt{15} }{25}$

Dividing both sides by 9:

$x = \dfrac{27 - 25\sqrt{15} }{225}$

Answer 2

$\large\bold{\underline{\underline{Given:-}}}$

$\sqrt{ \bigg(5^0 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

$\large\bold{\underline{\underline{To \: Find:-}}}$

The value of x

$\large\bold{\underline{\underline{Solution:-}}}$

$\sqrt{ \bigg(5^0 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

Rewriting 5⁰ = 1:

$\sqrt{ \bigg(1 + \dfrac{2}{3} \bigg) } = (0.6)^2 - 3x$

$\sqrt{ \dfrac{5}{3} }= (0.6)^2 - 3x$

Rationalise the LHS by multiplying √3/√3:

$\dfrac{\sqrt{15} }{3} = (0.6)^2 - 3x$

Rewriting (0.6)² as a fraction:

$\dfrac{\sqrt{15} }{3} = \bigg( \dfrac{6}{10} \bigg)^2 - 3x$

$\dfrac{\sqrt{15} }{3} = \dfrac{9}{25} - 3x$

Multiply both sides by 3:

$\sqrt{15} = \dfrac{27}{25} - 9x$

Find x:

$9x = \dfrac{27}{25} - \sqrt{15}$

$9x = \dfrac{27 - 25\sqrt{15} }{25}$

Dividing both sides by 9:

$x = \dfrac{27 - 25\sqrt{15} }{225}$