Factoring radicals: solve s = 32 + 4 sqrts of x by factoring

[onthesnap55]

Solve by factoring.
x=32+4sq roots of x

I don't know where to begin. I tried moving the x over, but this gave me a neg x and I have been lost from there.

 

[soroban]

Re: Factoring radicals

Hello, onthesnap55!

This takes a bit of imagination . . .

\(\displaystyle \text{Solve by factoring: }\;x \:=\:32 + 4\sqrt{x}\)

\(\displaystyle \text{We have: }\;x - 4\sqrt{x} - 32 \;=\;0\)

\(\displaystyle \text{Factor: }\;(\sqrt{x}-8)(\sqrt{x} + 4) \;=\;0\) .**


\(\displaystyle \text{And we have two equations to solve:}\)

. . \(\displaystyle \sqrt{x}-8\:=\:0\quad\Rightarrow\quad \sqrt{x} \:=\:8\quad\Rightarrow\quad x \:=\:64\)

. . \(\displaystyle \sqrt{x} + 4 \:=\:0 \quad\Rightarrow\quad \sqrt{x} \:=\:-4 \quad\Rightarrow\quad x \:=\:16\)

\(\displaystyle \text{We find that }x = 16\text{ doesn't check . . . It is an extraneous root.}\)

\(\displaystyle \text{Therefore, the only root is: }\;\boxed{x \:=\:64}\)


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** It can be factored with a substitution.

\(\displaystyle \text{We have: }\;x - 4\sqrt{x} - 32 \;=\;0\)

\(\displaystyle \text{Let }\,u \,=\,\sqrt{x} \quad\Rightarrow\quad u^2 \,=\,x\)

\(\displaystyle \text{Substitute: }\;u^2 - 4u - 32 \:=\:0\)

\(\displaystyle \text{Factor: }\;(u - 8)(u + 4) \:=\:0\)

. . \(\displaystyle \text{and we have: }\;\begin{Bmatrix}u - 8 &=& 0 & \Rightarrow & u &=& 8 \\ u + 4 &=&0 & \Rightarrow & u &=& \text{-}4 \end{Bmatrix}\)

\(\displaystyle \text{Back-substitute: }\;\begin{Bmatrix}\sqrt{x} &=& 8 & \Rightarrow & x&=&64 \\ \sqrt{x} &=&\text{-}4 & \Rightarrow & x &=&16 \end{Bmatrix}\quad\hdots \text{ See?}\)

 

[onthesnap55]

Re: Factoring radicals

Thank you, that was very helpfull. Can I ask another question?