Quadratic formula does not work

Question

If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get

$x = 1$ or $x = \frac{-4}{5}$

but the real zeros are: $x = -1$ or $x = \frac{4}{5}$

Can somebody explain me if the quadratic formula fails or me?

Answer 1

Probably it was meant to be $5x^2+x-4=0$. Then it would have those roots $-1$ and $\frac{4}{5}$.

Answer 2

$5x^2-x-4$ does indeed have roots $x = 1, -\frac{4}{5}$, so you did it correctly.

To illustrate:

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$=\frac{1 \pm \sqrt{1^2-4*5*(-4)}}{2(5)}$$ $$=\frac{1 \pm \sqrt{81}}{10}$$ $$= \frac{1\pm 9}{10}$$ $$x = 1, -\frac{4}{5}$$

As Peter said if you had meant $5x^2+x-4$ then your roots would have been $x = -1, +\frac{4}{5}$.

Answer 3

Given $5x^2 −x−4=0 $ you must use, $a=5, b=-1, c=-4$ in the formula$$\begin{align}x & =\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[2ex] & = \frac{-(-1)\pm\sqrt{(-1)^2-4(5)(-4)}}{2(5)} \\[2ex] & = \frac{1\pm\sqrt{1+80}}{10}\\[2ex] & = \frac{1\pm 9}{10}\\ \therefore x\in \{-\tfrac {4}{5}, 1\}\end{align}$$

Answer 4

The quadratic formula comes from "completing the square". Let's do that to your examples:

\begin{eqnarray*} 5x^2-x-4 &=& 0 \\ \\ x^2 - \tfrac{1}{5}x - \tfrac{4}{5} &=& 0 \\ \\ x^2 - \tfrac{1}{5}x &=& \tfrac{4}{5} \\ \\ \left(x - \tfrac{1}{10}\right)^{2} - \tfrac{1}{100} &=& \tfrac{4}{5} \\ \\ \left(x - \tfrac{1}{10}\right)^{2} &=& \tfrac{4}{5} + \tfrac{1}{100} \\ \\ \left(x - \tfrac{1}{10}\right)^{2} &=& \tfrac{81}{100} \\ \\ x - \tfrac{1}{10} &=& \pm \tfrac{9}{10} \\ \\ x &=& \tfrac{1}{10} \pm \tfrac{9}{10} \\ \\ x&=& -\tfrac{4}{5} \ \ \text{or} \ \ 1 \end{eqnarray*}

Answer 5

Plug in the roots the quadratic formula gave you

$5(1)^2 - (1) - 4 $

$=5 - 5 $

$= 0$

And your other root

$5(- \frac{4}{5})^2 - (-\frac{4}{5}) - 4 $

$ = \frac{20}{5} -\frac{(16+4)}{5}$

$= 0$

And you can verify your roots work.

The quadratic formula is correct.

I recommend looking at the proof of the quadratic formula if you never have before to possibly help convince you that it does indeed work and more importantly how it works.

if

$ax^2 + bx + c = 0$

then

$ax^2 + bx = -c$

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

$a(x^2 + \frac{b}{a}x + [\frac{b}{2a}]^2 - [\frac{b}{2a}]^2) = -c$

$a(x^2 + \frac{b}{a}x + [\frac{b}{2a}]^2) - a[\frac{b}{2a}]^2$ = -c

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

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

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

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

$x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The quadratic formula is devised by observing and analyzing the relationship between the objects $a$,$b$, and $c$ in the form of $ax^2 + bx + c = 0$.