Easiest Way To Solve A Radical Equation Using Substitution

Math problem: How to solve the radical equation with two radicals?

solve the radical equation cube root of (4 – x^2) + square root of (x^2 – 3) = 1

\begin{aligned}
\sqrt[3]{4-x^2}+\sqrt{x^2-3}=1
\end{aligned}
x=??

Soltion to the algebra math problem

Let substitute, k = √(x^2 – 3) then

\begin{aligned}
&k^2=x^2-3 \\
\\
\Rightarrow \ &k^2-1=x^2-4 \\
\\
\Rightarrow \ &1-k^2=4-x^2
\end{aligned}

Then

\begin{aligned}
&\sqrt[3]{4-x^2}+\sqrt{x^2-3}=1 \\
\\
\Rightarrow \ &\sqrt[3]{1-k^2}+k=1 \\
\\
\Rightarrow \ &\sqrt[3]{1-k^2}+k-1=0 \\
\\
\Rightarrow \ &\sqrt[3]{1-k^2}-(1-k)=0 \\
\\
\Rightarrow \ &\sqrt[3]{(1-k)(1+k)}-\sqrt[3]{(1-k)^3}=0 \\
\\
\Rightarrow \ &\sqrt[3]{1-k}(\sqrt[3]{1+k}-\sqrt[3]{(1-k)^2})=0 \\
\end{aligned}

So we get

\begin{aligned}
\sqrt[3]{1-k} = 0 \ \ \ or \ \ \ \sqrt[3]{1+k}-\sqrt[3]{(1-k)^2}=0
\end{aligned}

When

\begin{aligned}
& \sqrt[3]{1-k}=0\\
\\
\Rightarrow & \ 1-k=0 \\
\\
\Rightarrow & \ k=1 \\
\end{aligned}

When

\begin{aligned}
&\sqrt[3]{1+k}-\sqrt[3]{(1-k)^2}=0\\
\\
\Rightarrow \ &\sqrt[3]{1+k}=\sqrt[3]{(1-k)^2}\\
\\
\Rightarrow \ &1+k = (1-k)^2\\
\\
\Rightarrow \ &1+k = 1-2k+k^2\\
\\
\Rightarrow \ &3k = k^2\\
\\
\Rightarrow \ &3k - k^2 = 0\\
\\
\Rightarrow \ &k(3-k) = 0\\
\\
\Rightarrow \ &k= 0,3\\
\end{aligned}

We substitute k = √(x^2 – 3) so

\begin{aligned}
&k=\sqrt{x^2-3}\\
\\
\Rightarrow \ &k^2=x^2-3 \\
\\
\Rightarrow \ &x^2=3+k^2 \\
\\
\Rightarrow \ &x=\pm\sqrt{3+k^2}
\end{aligned}

When k = 1

\begin{aligned}
&x=\sqrt{3+k^2}\\
\\
\Rightarrow \ &x=\sqrt{3+1^2}\\
\\
\Rightarrow \ &x=\pm2\\
\end{aligned}

When k = 0

\begin{aligned}
&x=\sqrt{3+k^2}\\
\\
\Rightarrow \ &x=\sqrt{3+0^2}\\
\\
\Rightarrow \ &x=\pm \sqrt{3}\\
\end{aligned}

When k = 3

\begin{aligned}
&x=\sqrt{3+k^2}\\
\\
\Rightarrow \ &x=\sqrt{3+3^2}\\
\\
\Rightarrow \ &x=\pm \sqrt{12}\\
\\
\Rightarrow \ &x=\pm 2\sqrt{3}\\
\end{aligned}

So, the solutions to this radical equation are -2, 2, -√3, √3, -2√3 & 2√3

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