Question

Solve 3sqrt(x) = 8/(sqrt(9x - 32)) + sqrt(9x - 32)​

Answer 1

$\huge \color{gray}{ \tt{answer}}$

$x = 4$

Step-by-step explanation:

$3 \sqrt{x} = \frac{8}{(9x - 32)} + \sqrt{(9x - 32)}$

• move the expression to the left

$3 \sqrt{x} - \frac{8}{ \sqrt{9x - 32} } - \sqrt{9x - 32} = 0$

• transform the expression

$\frac{3 \sqrt{(9x - 32 x)} - 8 - \sqrt{9x - 32} }{ \sqrt{9x - 32} }$

• set the numerator

$3 \sqrt{(9x - 32)x} - 8 - (9x - 32) = 0$

• remove the parentheses

$3 \sqrt{9 {x}^{2} - 32x} + 24 - 9x = 0$

• move the expression to the right

$3 \sqrt{9 {x}^{2} - 32x } - 24 + 9x$

• divide both sides

$\sqrt{9 {x}^{2} - 32x } = - 8 + 3x$

• simplify the expression

$9 {x}^{2} - 32x = 9 {x}^{2} - 48x + 64$

• cancel equal terms

$- 32x = - 48x + 64$

• move the variable to the left

$- 32x = + 48x = 64$

• collect like terms

$16x = 64$

• divide both sides

$= \: \: x = 4$

• check the solution

$3 \sqrt{4} = \frac{8}{ \sqrt{9 \times 4 - 32} } + \sqrt{9 \times 4 - 32}$

• simplify

$6 = 6$

• so the solution is

$≈ \: \: x = 4$

$\huge \color {cyan}{ \tt{hope \: \: it \: \: helps}}$

Answer 2

Answer:

Step-by-step explanation:

$3\sqrt{x}=\frac{8}{\sqrt{9x-32}}+\sqrt{9x -32}\\\\\\3\sqrt{x}=\frac{8*\sqrt{9x-32}}{(\sqrt{9x-32})*(\sqrt{9x-32})}+\sqrt{9x-32}\\\\3\sqrt{x}=\frac{8*\sqrt{9x-32}}{9x-32}+\sqrt{9x-32}\\\\\\3\sqrt{x}=(\sqrt{9x-32})(\frac{8}{9x-32}+1)\\\\\frac{3\sqrt{x}}{\sqrt{9x-32}}=\frac{8+9x-32}{9x-32}\\\\\frac{3\sqrt{x}}{\sqrt{9x-32}}=\frac{9x - 24}{9x-32}\\\\Take \ square\\\\(\frac{3\sqrt{x}}{\sqrt{9x-32}})^{2}=(\frac{9x-24}{9x-32})^{2}\\\\\frac{9x}{9x-32}=\frac{(9x-24)^{2}}{(9x-32)^{2}}$

9x*(9x - 32) = (9x- 24)²

9x*9x - 9x*32 = 81x² - 2*9x*24 + 576

81x² - 288x = 81x² - 432x + 576

81x² - 288x - 81x² + 432x - 576 = 0

81x² - 81x² - 288x + 432x - 576 = 0

144x - 576 = 0

144x = 576

x = 576/144

x =4