Question

this question is on solving radicals for algebra 2; I also need help

Answer 1

Given:

Algebric equation is given

Required:

To calculate the value of x

Explanation:

$\begin{gathered} \sqrt{6-2x}\text{ +x=2\lparen x+15\rparen-9} \\ \\ \sqrt{6-2x\text{ }}\text{ +x=2x+21} \\ \\ \sqrt{6-2x}\text{ =x+21} \end{gathered}$$\begin{gathered} 6-2x=x^2+42x+441 \\ \\ x^2+44x+435=0 \\ \\ x=-15,\text{ x=-29} \end{gathered}$$\begin{gathered} when\text{ x=-15 } \\ \\ \sqrt{6-2(-15)}\text{ =-9} \\ \\ 2(-15+15)-9=-9 \\ \\ -9=-9\text{ so x=-15} \end{gathered}$$\begin{gathered} when\text{ x =-29} \\ \sqrt{6-2(-}29)+(-29)=35 \\ \\ =2(-29+15)-9=-37 \\ \\ 35\ne-37 \end{gathered}$

Required answer:

x=-15