How do you solve $x^{2} + 4 x + 1 = 0$ $x^2 + 4x + 1 = 0$ using the quadratic formula?

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How do you solve $x^{2} + 4 x + 1 = 0$ $x^2 + 4x + 1 = 0$ using the quadratic formula?

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Answer 1 · 2018-03-11T09:30:30

$x = - 2 \pm \sqrt{3}$ $x=-2+-sqrt3$

Explanation:

$Given the quadratic equation in standard form$ $"Given the quadratic equation in standard form"$

$∙ x a x^{2} + b x + c = 0 x; a \neq 0$ $•color(white)(x)ax^2+bx+c=0 color(white)(x);a!=0$

$We can solve for x using the quadratic formula$ $"We can solve for x using the "color(blue)"quadratic formula"$

$∙ x x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)$

$x^{2} + 4 x + 1 = 0 is in standard form$ $x^2+4x+1=0" is in standard form"$

$with a = 1, b = 4 and c = 1$ $"with "a=1,b=4" and "c=1$

$\Rightarrow x = \frac{- 4 \pm \sqrt{16 - 4}}{2} = \frac{- 4 \pm \sqrt{12}}{2}$ $rArrx=(-4+-sqrt(16-4))/2=(-4+-sqrt12)/2$

$x = \frac{- 4 \pm 2 \sqrt{3}}{2} = - 2 \pm \sqrt{3} \leftarrow exact values$ $color(white)(x)=(-4+-2sqrt3)/2=-2+-sqrt3larrcolor(red)"exact values"$

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How do you solve $x^{2} + 4 x + 1 = 0$ $x^2 + 4x + 1 = 0$ using the quadratic formula?

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