Question #ae517

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Question #ae517

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Answer 1 · 2016-10-24T22:30:35

$x = \pm 2 \sqrt{7}$ $x=+-2sqrt(7)$

Explanation:

$f (x) = g (x)$ $f(x) = g(x)$

$\Rightarrow x + 5 = \frac{3}{x - 5}$ $=> x+5 = 3/(x-5)$

$\Rightarrow (x + 5) (x - 5) = 3$ $=> (x+5)(x-5) = 3$

$\Rightarrow x^{2} - 25 = 3$ $=> x^2 - 25 = 3$

At this point it would be simplest to just add $25$ $25$ to both sides and take a square root, but as the question refers to the quadratic formula, we will instead put it in standard form and proceed in that manner.

$\Rightarrow x^{2} - 28 = 0$ $=> x^2 - 28 = 0$

$\Rightarrow 1 x^{2} + 0 x - 28 = 0$ $=> 1x^2 + 0x - 28 = 0$

$\Rightarrow x = \frac{- 0 \pm \sqrt{0^{2} - 4 (1) (- 28)}}{2 (1)}$ $=> x = (-0+-sqrt(0^2-4(1)(-28)))/(2(1))$

$= \frac{\pm 4 \sqrt{7}}{2}$ $=(+-4sqrt(7))/2$

$= \pm 2 \sqrt{7}$ $=+-2sqrt(7)$

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Question #ae517

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