How do you solve $x^{2} + 4 x + 2 = 0$ $x^2+4x+2=0$ ?

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How do you solve $x^{2} + 4 x + 2 = 0$ $x^2+4x+2=0$ ?

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Answer 1 · 2015-06-30T11:24:57

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

Explanation:

For any equation of the form
$XXXX$ $color(white)("XXXX")$ $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
the solutions can be obtained using the quadratic formula:
$XXXX$ $color(white)("XXXX")$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$
(this formula shows up often enough that it is worth memorizing)

Substituting the values from $x^{2} + 4 x + 2 = 0$ $x^2+4x+2 = 0$ for $a, b, and c$ $a,b, and c$
$XXXX$ $color(white)("XXXX")$ $x = \frac{- 4 \pm \sqrt{4^{2} - 4 (1) (2)}}{2 (1)}$ $x= (-4+-sqrt(4^2-4(1)(2)))/(2(1))$

$XXXX$ $color(white)("XXXX")$ $x = \frac{- 4 \pm \sqrt{8}}{2}$ $x = (-4+-sqrt(8))/2$

$XXXX$ $color(white)("XXXX")$ $x = - 2 \pm \sqrt{2}$ $x=-2+-sqrt(2)$

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How do you solve $x^{2} + 4 x + 2 = 0$ $x^2+4x+2=0$ ?

1 Answer

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How do you solve $x^{2} + 4 x + 2 = 0$ $x^2+4x+2=0$ ?