How Do I Find The Solutions For The Quadratic Equation : X^2-2x-8 =0 Step By Step

The Quadratic Equation: $$x^2-2x-8=0$$

The solution for $$x^2+px+q=0$$ is:

$$x_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}$$

 

here: $$p=-2$$ and $$q=-8$$

therefor: $$x_{1,2}=-\frac{-2}{2}\pm\sqrt{\frac{(-2)^2}{4}-(-8)}$$

$$x_{1,2}=1\pm\sqrt{\frac{4}{4}+8}$$

$$x_{1,2}=1\pm\sqrt{1+8}$$

$$x_{1,2}=1\pm\sqrt{9}$$

$$x_{1,2}=1\pm3$$

$$x_1=1+3=4$$

$$x_2=1-3=-2$$

$$\text{so}\quad x_1=4 \quad \text{and} \quad x_2=-2$$

Bye

 

You could solve it using the quadratic formula.

https://www.youtube.com/watch?v=O8ezDEk3qCg

but it is easy enough to do by factorising.

$$x^2-2x-8=0$$

You need to find two numbers  that multiply to give -8 (straight away you know that one is neg and one is pos)

and they have to add to -2 (so the bigger looking one is neg)

-4 and 2 That was easy.

$$(x-4)(x+2)=0$$

Now, if you are happy you can give me a thumbs up.  I would like that!

Now, if 2 things are going to multiply to give 0 then one (or both) of them must be zero.

so  x-4=0   or  x+2=0

x=4   or   x=-2