Find the solution: 6=\sqrt{x^2-2x+12}

abondantQ 2021-09-22 Answered
Find the solution:
\(\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}\)

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Expert Answer

Jayden-James Duffy
Answered 2021-09-23 Author has 17064 answers

Radical or Exponents or Power is defined as the number of times that a variable or constant is being repeated. It is also called power raised. Exponents are used to describing number of times in a single notation
The radical equation given to us
\(\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}\)
To find the solution for the given radical equation
Solution to the question
We have,
\(\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}\)
Squaring both sides we obtain,
\(\displaystyle{6}^{{2}}={\left(\sqrt{{{x}^{{2}}-{2}{x}+{12}}}\right)}^{{2}}\)
\(\displaystyle{36}={x}^{{2}}-{2}{x}+{12}\)
\(\displaystyle{x}^{{2}}-{2}{x}+{12}-{36}={0}\)
\(\displaystyle{x}^{{2}}-{2}{x}-{24}={0}\)
\(\displaystyle{x}^{{2}}-{6}{x}+{4}{x}={0}\)
\(\displaystyle{x}{\left({x}-{6}\right)}+{4}{\left({x}-{6}\right)}={0}\)
\(\displaystyle{\left({x}+{4}\right)}{\left({x}-{6}\right)}={0}\)
either \(x=-4\ \text{or}\ x=6\)
There are two valid solutions \(\displaystyle{x}=-{4}{\quad\text{and}\quad}{x}={6}\)

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