How do you solve $2 x^{2} - 4 x - 5 = 0$ $2x^2 - 4x - 5=0$ using the quadratic formula?

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How do you solve $2 x^{2} - 4 x - 5 = 0$ $2x^2 - 4x - 5=0$ using the quadratic formula?

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Answer 1 · 2017-01-18T00:05:51

$x = 1 \pm \frac{\sqrt{14}}{2}$ $x = 1+-sqrt(14)/2$

Explanation:

$2 x^{2} - 4 x - 5 = 0$ $2x^2-4x-5 = 0$

is of the form:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

with $a = 2$ $a=2$ , $b = - 4$ $b=-4$ and $c = - 5$ $c=-5$ .

This has roots given by the quadratic formula:

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

$x = \frac{4 \pm \sqrt{{(- 4)}^{2} - 4 (2) (- 5)}}{2 (2)}$ $color(white)(x) = (4+-sqrt((-4)^2-4(2)(-5)))/(2(2))$

$x = \frac{4 \pm \sqrt{16 + 40}}{4}$ $color(white)(x) = (4+-sqrt(16+40))/4$

$x = \frac{4 \pm \sqrt{56}}{4}$ $color(white)(x) = (4+-sqrt(56))/4$

$x = \frac{4 \pm \sqrt{2^{2} \cdot 14}}{4}$ $color(white)(x) = (4+-sqrt(2^2*14))/4$

$x = \frac{4 \pm 2 \sqrt{14}}{4}$ $color(white)(x) = (4+-2sqrt(14))/4$

$x = 1 \pm \frac{\sqrt{14}}{2}$ $color(white)(x) = 1+-sqrt(14)/2$

$color(white)()$
Footnote

The quadratic formula is very useful, but is it just a "magical" formula to you, or do you know how to derive it?

Here's one way:

Given:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

We find:

$0 = \frac{1}{a} (a x^{2} + b x + c)$ $0 = 1/a(ax^2+bx+c)$

$0 = x^{2} + \frac{b}{a} x + \frac{c}{a}$ $color(white)(0) = x^2+b/ax+c/a$

$0 = x^{2} + 2 \frac{b}{2 a} x + \frac{b^{2}}{{(2 a)}^{2}} - \frac{b^{2}}{{(2 a)}^{2}} + \frac{c}{a}$ $color(white)(0) = x^2+2b/(2a)x+b^2/(2a)^2-b^2/(2a)^2+c/a$

$0 = {(x + \frac{b}{2 a})}^{2} - \frac{b^{2} - 4 a c}{4 a^{2}}$ $color(white)(0) = (x+b/(2a))^2-(b^2-4ac)/(4a^2)$

Add $\frac{b^{2} - 4 a c}{4 a^{2}}$ $(b^2-4ac)/(4a^2)$ to both ends and transpose to get:

${(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$ $(x+b/(2a))^2 = (b^2-4ac)/(4a^2)$

Take the square root of both sides, allowing for both positive and negative square roots to find:

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

Subtract $\frac{b}{2 a}$ $b/(2a)$ from both ends to find:

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

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How do you solve $2 x^{2} - 4 x - 5 = 0$ $2x^2 - 4x - 5=0$ using the quadratic formula?

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