Solving a quadratic equation

Question

amtran_bus · Answer

attachment

amtran_bus · Answer

Can you straight up use the quadratic formula?

inkyvoyd · Answer

One can ALWAYS straight up use the quadratic formula, but it is not always the fastest.

campbell_st · Answer

I think you need to general quadratic formula.

amtran_bus · Answer

Can you guide me through the steps?

inkyvoyd · Answer

I can show the answer by factoring.

mertsj · Answer

a=2, b=4, c=-3

amtran_bus · Answer

$$-4\pm \sqrt{16-4(2)(-3)}$$

amtran_bus · Answer

all over 4...

inkyvoyd · Answer

(2x )(x )
note how we have -3 and +4x, so (2x-1)(x+3)=2x^2+4x-3

inkyvoyd · Answer

that wasn't too hard in terms of factoring - always try factoring before using the quadratic equation.

amtran_bus · Answer

Thanks inkyvoyd. The answer is the attachment below, will I get that particular answer?

mertsj · Answer

IT DOES NOT FACTOR

amtran_bus · Answer

:0

inkyvoyd · Answer

And oftentimes you incorrectly factor - (2x-1)(x+3)=2x^2-5x-3

inkyvoyd · Answer

Anyways, easy way to tell that it won't factor is if b^2-4ac (the discriminant) is not a perfect square, since it must be a perfect square to factor and have a rational answer.

amtran_bus · Answer

how can I get this answer?

inkyvoyd · Answer

ewh webassign, just use the formula -

$\Huge x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
a=2,b=4,c=-3
plug and chug

amtran_bus · Answer

$$-4\pm \sqrt{16-4(2)(-3)}$$ over 4?

inkyvoyd · Answer

yup.

amtran_bus · Answer

But I get -4 plus or minus sqrt if 40 over 4

amtran_bus · Answer

$$-4\pm \sqrt{40}$$ over 4

mertsj · Answer

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