Question

Solve please thanks x+5= the square root of 7x+53

Answer 1

$x+5= \sqrt{7x+53} \ \ \ and\ \ \ 7x+53 \geq 0\ \ \ \Rightarrow\ \ \ D=\bigg{\langle}- 7\frac{\big{4}}{\big{7}} ,+\infty\bigg{)}\\\\(x+5)^2=( \sqrt{7x+53})^2\\\\x^2+10x+25=7x+53\\x^2+3x-28=0 \\x^2+4x-7x-28=0\\x(x+4)-7(x+4)=0\\(x+4)(x-7)=0\ \ \ \Leftrightarrow\ \ \ x+4=0\ \ \ or\ \ \ x-7=0\\\\x=-4\ \in D\ \ \ and\ \ \ x=7\ \in D\\\\Ans.\ x=-4\ \ or\ \ x=7$

Answer 2

$x+5= \sqrt{7x+53} \\ \\ (x+5)^2= (\sqrt{7x+53} )^2 \\ \\ x^2+10x+25=7x+53 \\ \\ x^2+3x-28=0 \\ \\ (x-4)(x+7)=0 \\ \\ x-4=0 \\ x=4 \\ \\ x+7=0 \\ x=-7 \\ \\ x=-7,4$