Question

I have this math problem I've​ been struggling with, it has multiple parts but it's basically the same idea. the question is solve the problems. With each step, please.

Find the real solutions of each equation. (x + 2)4/3-5(x + 2)2/3 + 6-0 3x3 + 4x2-27x + 36

Answer 1

Solution: 1)

   $\left ( x+2 \right )^{}\frac{4}{3}-5\left ( x+2 \right )^{\frac{2}{3}}+6=0$

   $\therefore[ \left ( x+2 \right )^{2}]^{}\frac{2}{3}-5\left ( x+2 \right )^{\frac{2}{3}}+6=0$

put _{$\left ( x+2 \right )^{\frac{2}{3}}=u$}

$\therefore u^{^{2}}-5u+6=0$

$\therefore \left ( u-3 \right )\left ( u-2 \right )=0$

$\therefore u=3\: \: or\: \: \: u=2$

Resubstitute the value of $\left ( x+2 \right )^{\frac{2}{3}}$

$\therefore \left ( x+2 \right )^{\frac{2}{3}}=3\: \: \: or\: \: \left ( x+2 \right )^{\frac{2}{3}}=2$

$\therefore x+2=3^{\frac{3}{2}}\: \: \: or\: \: \: \: x+2=2^{\frac{3}{2}}$

$\therefore x=3^{\frac{3}{2}}-2\: \: \: or\: \: \: \: x=2^{\frac{3}{2}}-2$

2)    $3x^{3}+4x^{2}-27x-36=0$

By synthetic division

3	3	4	-27	-36
		9	39	36
	3	13	12	0

$\therefore \left ( x-3 \right )\left ( 3x^{2} \right +13x+12)=0$

$\therefore \left ( x-3 \right )\left ( x+3 \right )\left ( 3x+4 \right )=0$

$\therefore \left ( x^{2}-9 \right )\left ( 3x+4 \right )=0$

$\therefore x=\pm 3\: \: \: or\: \: \: x=-\frac{4}{3}$

3) $2x-1=-\sqrt{2-x}$

squaring on both side we get

$\left ( 2x-1 \right )^{2}=2-x$

$\therefore 4x^{2}-4x+1=2-x$

$\therefore 4x^{2}-3x-1=0$

$\therefore \left ( x-1)\left ( 4x+1 \right ) \right =0$

$\therefore x=1\: \: \: or\: \: \: x=-\frac{1}{4}$

4) $\sqrt{2x-5}-\sqrt{x-3}=1$

Sqquaring on both side

$\therefore 2x-5+x-3-2\sqrt{2x-5}\sqrt{x-3}=1$

$\therefore 3x-9=2\sqrt{2x-5}\sqrt{x-3}$

Squaring on both side

$\therefore 9x^{2}-54x+81=8x^{2}-44x+60$

$\therefore x^{2}-10x+21=0$

$\therefore \left ( x-7 \right )\left ( x-3 \right )=0$

$\therefore x=7\: \: \: or\: \: \: x=3$

5)    $\sqrt[4]{5x^{2}-6}=x$

$5x^{2}-6=x^{4}$

$\therefore x^{4}-5x^{2}+6=0$

$\therefore \left ( x^{2}-2 \right )\left ( x^{2} -3\right )=0$

$x=\pm \sqrt{2}\: \: \: or\: \: \: \: x=\pm \sqrt{3}$