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How do you solve the equation $3 {(x - 3)}^{2} + 2 = 26$ $3(x-3)^2+2=26$ ?

1 Answer

Sep 26, 2017

$x = 3 \pm 2 \sqrt{2}$ $x=3+-2sqrt(2)$

Explanation:

$3 {(x - 3)}^{2} = 26$ $3(x-3)^2=26$

$\to 3 {(x - 3)}^{2} = 24$ $rarr 3(x-3)^2=24$

$\to {(x - 3)}^{2} = 8$ $rarr (x-3)^2=8$

$\to x^{2} - 6 x + 9 = 8$ $rarr x^2-6x+9=8$

$\to x^{2} - 6 x + 1 = 0$ $rarr x^2-6x+1=0$

Applying the quadratic formula
$XXX x = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (1) (1)}}{2 (1)}$ $color(white)("XXX")x=(-(-6)+-sqrt((-6)^2-4(1)(1)))/(2(1))$

$XxXX = \frac{6 \pm \sqrt{32}}{2}$ $color(white)("XxXX")=(6+-sqrt(32))/2$

$XxXX = \frac{6 \pm 4 \sqrt{2}}{2}$ $color(white)("XxXX")=(6+-4sqrt(2))/2$

$XxXX = 3 \pm 2 \sqrt{2}$ $color(white)("XxXX")=3+-2sqrt(2)$

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