How do you solve ${(x - 3)}^{2} = 5$ $(x-3)^2 = 5$ using the quadratic formula?

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How do you solve ${(x - 3)}^{2} = 5$ $(x-3)^2 = 5$ using the quadratic formula?

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Answer 1 · 2018-05-09T13:35:06

${(x - 3)}^{2} = 5$ $(x-3)^2=5$

$x_{1} = 3 + \sqrt{5}$ $x_1=3+\sqrt(5)$
$x_{2} = 3 - \sqrt{5}$ $x_2=3 -\sqrt(5)$

Explanation:

${(x - 3)}^{2} = 5$ $(x-3)^2=5$
$x^{2} - 6 \cdot x + 9 = 5$ $x^2-6*x+9=5$
$x^{2} - 6 \cdot x + 9 - 5 = 0$ $x^2-6*x+9-5=0$
$x^{2} - 6 \cdot x + 4 = 0$ $x^2-6*x+4=0$

$a \cdot x^{2} + b \cdot x + c = 0$ $a*x^2+b*x+c=0$

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$ $x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)$

$a = 1$ $a=1$
$b = 6$ $b=6$
$c = 4$ $c=4$

$x_{1, 2} = \frac{- (- 6) \pm \sqrt{6^{2} - 4 \cdot 1 \cdot 4}}{2 \cdot 1} =$ $x_(1,2)=(-(-6) +- \sqrt(6^2-4*1*4))/(2*1)=$
$= \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1} =$ $=(6 +- \sqrt(36-16))/(2*1)=$
$= \frac{6 \pm \sqrt{20}}{2} =$ $=(6 +- \sqrt(20))/(2)=$
$= \frac{6 \pm \sqrt{4 \cdot 5}}{2} =$ $=(6 +- \sqrt(4*5))/(2)=$
$= \frac{6 \pm 2 \cdot \sqrt{5}}{2} =$ $=(6 +- 2*\sqrt(5))/(2)=$
$= 3 \pm \sqrt{5}$ $=3 +-\sqrt(5)$

$x_{1} = 3 + \sqrt{5}$ $x_1=3+\sqrt(5)$
$x_{2} = 3 - \sqrt{5}$ $x_2=3 -\sqrt(5)$

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How do you solve ${(x - 3)}^{2} = 5$ $(x-3)^2 = 5$ using the quadratic formula?

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How do you solve (x−3)2=5(x-3)^2 = 5 using the quadratic formula?

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How do you solve ${(x - 3)}^{2} = 5$ $(x-3)^2 = 5$ using the quadratic formula?