Algebra

Find the smallest value of x in (x^2 + 10x + 25 = 64).

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\(x^2 + 10x + 25 = 64\\ x^2+10x-39=0\)

          p          q

\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=-\frac{10}{2}\pm \sqrt{(\frac{10}{2})^2+39}\\ x=-5\pm 8\\ x\in\{-13,3\}\)

The smallest value of x in (x^2 + 10x + 25 = 64) is - 13.

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There is another way to solve this problem. Notice that the left side is the square of a binomial.

\((x + 5)^2 = 64\)

\(x + 5 = 8 \qquad x + 5 = -8\)

x = 3 or -13.