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# Algebra

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Find the smallest value of x such that x^2 + 10x + 25 = 64.

Feb 14, 2022

#1
+13581
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Find the smallest value of x in (x^2 + 10x + 25 = 64).

Hello Guest!

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

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.

!

Feb 14, 2022
edited by asinus  Feb 14, 2022
#2
+1209
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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.

Feb 14, 2022