Question

what is 2x+square root x+1=8 solve for x

Answer 1

$\bf 2x+\sqrt{x+1}=8\implies \sqrt{x+1}=8-2x\quad \leftarrow \textit{now we square both sides} \\\\\\ (\sqrt{x+1})^2=(8-2x)^2\implies x+1=8^2-32x+(2x)^2 \\\\\\ x+1=64-32x+2^2x^2\implies x+1=64-32x+4x^2 \\\\\\ 0=4x^2-32x-x+64-1\implies 0=4x^2-33x+63 \\\\\\ 0=(4x-21)(x-3)\implies \begin{cases} 0=4x-21\\ 21=4x\\\\ \boxed{\frac{21}{4}=x}\\ --------\\ 0=x-3\\ \boxed{3=x} \end{cases}$

Answer 2

Hello,

On my keybord i have no touch "square root (√)" but "V" yes .

$2x+ \sqrt{x+1} =8\\ \sqrt{x+1} =8-2x\\ x+1=(8-2x)^2\\ x+1=64-32x+4x^2\\ 4x^2-33x+63=0\\ x= \frac{21}{4} \ or\ x=3$

Verification:
if x=3 then 2*3+√(3+1)=6+2=8

if x=21/4 then 13≠8 is to exclude.