# 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**