What is the solution set of the equation $x^{2} + \sqrt{2} x = \frac{1}{2}$ $x^2+sqrt2x=1/2$ ?

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What is the solution set of the equation $x^{2} + \sqrt{2} x = \frac{1}{2}$ $x^2+sqrt2x=1/2$ ?

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Answer 1 · 2017-07-10T06:53:35

$x = - \frac{1}{\sqrt{2}} + 1$ $x=-1/sqrt2+1$ or $- \frac{1}{\sqrt{2}} - 1$ $-1/sqrt2-1$

Explanation:

$x^{2} + \sqrt{2} x = \frac{1}{2}$ $x^2+sqrt2x=1/2$ can be written as

$2 x^{2} + 2 \sqrt{2} x - 1 = 0$ $2x^2+2sqrt2x-1=0$

Hence using quadratic formula

$x = \frac{- 2 \sqrt{2} \pm \sqrt{{(2 \sqrt{2})}^{2} - 4 \times 2 \times (- 1)}}{2 \times 2}$ $x=(-2sqrt2+-sqrt((2sqrt2)^2-4xx2xx(-1)))/(2xx2)$

$= \frac{- 2 \sqrt{2} \pm \sqrt{8 + 8}}{4}$ $=(-2sqrt2+-sqrt(8+8))/4$

$= - \frac{\sqrt{2}}{2} \pm \frac{4}{4}$ $=-sqrt2/2+-4/4$

$= - \frac{1}{\sqrt{2}} \pm 1$ $=-1/sqrt2+-1$

i.e $- \frac{1}{\sqrt{2}} + 1$ $-1/sqrt2+1$ or $- \frac{1}{\sqrt{2}} - 1$ $-1/sqrt2-1$

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What is the solution set of the equation $x^{2} + \sqrt{2} x = \frac{1}{2}$ $x^2+sqrt2x=1/2$ ?

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Explanation:

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