Solve each radical equation. Check potential solutions. \sqrt{5x}=x-10

Harlen Pritchard 2021-09-29 Answered
Solve each radical equation. Check potential solutions.
\(\displaystyle\sqrt{{{5}{x}}}={x}-{10}\)

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Expert Answer

tabuordg
Answered 2021-09-30 Author has 7367 answers

Step 1
To solve the radical equation, isolate the radical expression and taking square on both sides to solve the equation.
Now the equation becomes quadratic after squaring each side, then solve the quadratic equation by factoring the terms.
To check the potential solution, substitute back the two values of x in the radical equation to check which value of x satisfying the equation.
Step 2
Isolate the radical expression from the equation,
\(\displaystyle\sqrt{{{5}{x}}}={x}-{10}\)
Squaring each side to solve the equation,
\(\displaystyle{5}{x}={\left({x}-{10}\right)}^{{{2}}}{5}{x}={x}^{{{2}}}+{100}={20}{x}\)
\(\displaystyle{x}^{{{2}}}-{20}{x}-{5}{x}+{100}={0}\)
\(\displaystyle{x}^{{{2}}}-{25}{x}+{100}={0}\)
\(\displaystyle{x}^{{{2}}}-{20}{x}-{5}{x}+{100}={0}\)
\(x(x-20)-5(x-20)=0\)
\(x=5, 20\)
Now, checking these two values of x by substituting back in the radical equation,
Substituting x=5 in the left hand side of the radical equation to check,
\(\displaystyle\sqrt{{{5}{x}}}+{10}\)
\(\displaystyle=\sqrt{{{5}\times{5}}}+{10}\)
\(\displaystyle=\sqrt{{{25}}}+{10}\)
\(=5+10\)
\(=15\)
Since the right hand side of radical equation is x=5, which is not equal to L.H.S which is 15, so x=5 is not the solution of this equation.
Substituting x=20 in the equation to check,
\(\displaystyle\sqrt{{{5}{x}}}+{10}\)
\(\displaystyle=\sqrt{{{5}\times{20}}}+{10}\)
\(\displaystyle=\sqrt{{{100}}}+{10}\)
\(=10+10\)
\(=20\)
Since the R.H.S of equation is x which is 20, so L.H.S is equal to the R.H.S, so x=20 is the solution of the radical equation.

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