Respuesta :
[tex]\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}[/tex]
Hello,
On my keybord i have no touch "square root (√)" but "V" yes .
[tex]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\\ [/tex]
Verification:
if x=3 then 2*3+√(3+1)=6+2=8
if x=21/4 then 13≠8 is to exclude.
On my keybord i have no touch "square root (√)" but "V" yes .
[tex]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\\ [/tex]
Verification:
if x=3 then 2*3+√(3+1)=6+2=8
if x=21/4 then 13≠8 is to exclude.