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{x}^(2) + 5 = 21what are the roots x=x=

1 Answer

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We have the quadratic equation:


x^2+5=21

We have to find the roots.

Then, we rearrange the equation as:


\begin{gathered} x^2+5=21 \\ x^2+5-21=0 \\ x^2-16=0 \end{gathered}

As 16 is the square of 4, we have a difference of squares that can be written as:


\begin{gathered} x^2-16=0 \\ x^2-4^2=0 \\ (x-4)(x+4)=0 \end{gathered}

Then, we have the function factorized and the roots are x=4 and x=-4.

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