Question

Solve using the Quadratic Formula.

2x² - 6 = 0

Answer 1

Do you really need to? You can simply add 6 to both sides divide by 2 and then square root both sides.

Answer 2

yes I really need to

Answer 3

that sucks well can you tell me what a b and c are?

Answer 4

\[\large \begin{align}
2x^2 - 6 &= 0\\
2x^2&=6\\
x^2&=6\div2 \\
x^2&=3\\
x&=\pm\sqrt{3}
\end{align}\]

Answer 5

I see no need to use the quadratic formula here :$

Answer 6

me too but if the questions asks you to do it then you have no choice

Answer 7

why do I not need to use it.

Answer 8

because you can solve it algebraically

Answer 9

can you tell me what a b and c are? I know you have done problems like this.

Answer 10

I'm so bad at math but here I go x^2-2x-6 so b=2,c=6

Answer 11

A wild -2x appeared, the leading coefficient disappeared..

Answer 12

\(\large x^2-2x-6=0\)

A should be 1
B should be -2
C should be -6

Answer 13

ok

Answer 14

A quadratic equation has the form \(\large ax^2+bx+c=0\), where a, b and c are some constants.
Your question is \(\large x^2−2x−6=0\), we can rewrite that as \(\large (1)x^2+(-2)x+(-6)=0\)

Answer 15

That's why a=1; b=-2 and c=-6

Answer 16

Notice the +/- sign, they are very important

Now use the quadratic formula, which is \(\huge x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), simply plug in the values (BE CAREFUL WITH SIGNS!! I would suggest you to put parenthesis)
\(\huge x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-6)}}{2(1)}\)
And the rest is arithmetic :P

Answer 17

So the first difficult part is to identify what's a,b and c (just be careful with the signs and you'll be fine) then simply plug in the values in and you're done!

Answer 18

Thank you @zepp

Answer 19

\(\huge x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-6)}}{2(1)}\\\huge \frac{2\pm\sqrt{4+24}}{2}=\frac{2\pm\sqrt{28}}{2}\)
There you :D

Answer 20

You are welcome :)

Answer 21

If it says "use the quadratic formula" you have to use it. If you use an alternative method, you will get partial credit. If you receive a warning from the teacher about it and continue to do it, you will lose more points. If you use a method other than the one asked of you on a test, you will receive zero points.

Answer 22

^