Question

Fubd tge exact solutions (roots) using the quadratic forumula.

1) 4x^2 + 3x - 2 = 0

2) 2x^2 + 5x = 9

Medal will be awarded to those who can solve it and explain HOW! Thanks in advance!

Answer 1

You are given a quadratic in the form ax+bx+c = 0, first step is to see if the quadratic can be factored and if it cannot, we use the quadratic formula which is -b +- sqrt(b^2-4ac)/2a

Answer 2

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

Answer 3

For the first one right?

Answer 4

You will have a positive solution and a negative solution also known as the roots

Answer 5

yes

Answer 6

For the second one, you will use the same formula and use 0 for c

Answer 7

Wouldn't you subtract 9 to the otherside so c = -9?

Answer 8

Which one are we talking about?

Answer 9

Number 2, don't you need it to equal 0, so subtract 9?

Answer 10

For Number 1, I got 3 +- √41 / 8
What do I do from there?

Answer 11

Yes, so it would be 2x^2+5x-9 = 0

Answer 12

You're right

Answer 13

quadratic equation is set to 0

Answer 14

You should get one positive and one negative solution

Answer 15

For Number 1, I got 3 +- √41 / 8
What do I do from there?
Not sure how to proceed or what answer will look like.

Answer 16

I got two decimal approximations

Answer 17

So one answer would be 3.4 and the other would be -1.2

Answer 18

How do I break up Sqrt 41?

Answer 19

it should be set up like this: -3 +- sqrt(3^2-4(4)(-2)/2(4)

Answer 20

for the first one

Answer 21

Thats what I got.

Answer 22

I simplified down to I got 3 +- √41 / 8

Answer 23

What nxt?

Answer 24

Got the answer, I wasn't sure if i was allowed to use calculator on sqrt41

Answer 25

I have been allowed to use the calculator on quadratics

Answer 26

You would solve what is under the radical and then do the division making sure you get two solutions

Answer 27

Ah, i got different answers..