Question

What are the beginning steps of solving this?3D%208" id="TexFormula1" title="2x + \sqrt{x + 1} = 8" alt="2x + \sqrt{x + 1} = 8" align="absmiddle" class="latex-formula">

Answer 1

Isolate the radical term, then square both sides:

$2x+\sqrt{x+1}=8$

$\sqrt{x+1}=8-2x$

Note that $\sqrt{x+1}$ is non-negative and requires that $x\ge-1$ in order to be defined, while $8-2x$ when $x>4$. This means we can only find real solutions in the range $-1\le x\le4$.

$(\sqrt{x+1})^2=(8-2x)^2$

$x+1=(8-2x)^2$

Now just expand the RHS and simplify as much as possible:

$x+1=64-32x+4x^2$

$4x^2-33x+63=0$

$(x-3)(4x-21)=0$

$\implies x=3,x=\dfrac{21}4$

Now, $\dfrac{21}4=5.25$, which is larger than 4, so we omit that solution. Then $x=3$ is the only solution.