Question

one more question please help :)

Answer 1

were is that question

Answer 2

the screenshot

Answer 3

\[\large\rm x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{3\pm\sqrt{9+16}}{2}\]

Answer 4

If you look closely at the denominators,
you might be able to easily figure out your \(\large\rm a\) value.

Answer 5

oh that is correct

Answer 6

the a value would be a right?

Answer 7

We're looking for a number...

Answer 8

no find out it

Answer 9

They both have a 2,
so \(\large\rm a\) is not the 2.
What other invisible number down there could it be?

Answer 10

\(\large\rm 2a=2\)
\(\large\rm ~~a=~?\)

Answer 11

would it just be 0 then?

Answer 12

Hmm maybe. Let's think about it though.
If a=0,

Then 2a=0.
Oops that didn't give us 2 like we wanted :(

Answer 13

so it isnt 2?

Answer 14

Try to remember this,
whenever you look at a number... there is always an invisible 1 multiplying it :)
You can use it any time that it's helpful, like right now,\[\large\rm 2(a)=2(1)\]Do you see the a value now?

Answer 15

yes i do

Answer 16

lol what is it? :)

Answer 17

well since there is a an invisible 1 , 2(1)= 2

Answer 18

is that what you ment ? for the denomimator..

Answer 19

You still haven't told me what \(\large\rm a\) is.

Answer 20

a=?

Answer 21

If you prefer, you can solve it algebraically.
2a=2,
divide each side by 2.

Answer 22

oh sorry 1

Answer 23

\[\large\rm \frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{3\pm\sqrt{9+16}}{2}\]Ok good.
Let's plug that in anywhere else in the formula that we have an a,\[\Large\rm \frac{-b\pm\sqrt{b^2-4(1)c}}{2(1)}=\frac{3\pm\sqrt{9+16}}{2}\]

Answer 24

So we've matched up our denominators, good.

Answer 25

Let's try to match up these pieces in front,\[\Large\rm \frac{\color{orangered}{-b}\pm\sqrt{b^2-4(1)c}}{2(1)}=\frac{\color{orangered}{3}\pm\sqrt{9+16}}{2}\]These also have to be the same.
So we have: \(\large\rm \color{orangered}{-b=3}\)

Answer 26

ok

Answer 27

-b=3,
solve for b :)

Answer 28

-3?

Answer 29

Ok great, b=-3.
Let's plug it in.\[\large\rm \frac{-(-3)\pm\sqrt{(-3)^2-4(1)c}}{2(1)}=\frac{3\pm\sqrt{9+16}}{2}\]What are we missing? Ah, the c value!\[\large\rm \frac{-(-3)\pm\sqrt{(-3)^2\color{orangered}{-4(1)c}}}{2(1)}=\frac{3\pm\sqrt{9\color{orangered}{+16}}}{2}\]These must be equal as well.

Answer 30

-4(1)c=16

Solve for c.

Answer 31

-4

Answer 32

a=1
b=-3
c=-4

Ok great!

Remember to write your quadratic in standard form?
\(\large\rm ax^2+bx+c\)
Plug in the values :)

Answer 33

awesome thanks!!

Answer 34

np