Question

PLEASE HELP!!! REALLY QUICK!
What are the exact solutions of x2 - x - 4 = 0?

Answer 1

I believe D is the correct answer

Answer 2

Well, do you remember what the formula for finding the two solutions of a quadratic equation looked like ?

Answer 3

Given a general quadratic form, say:

a*x^2 + b*x + c = 0, where a,b,c are real constants.

Answer 4

yes it is\[x = -b \sqrt{b^2 - 4ac} \over 2a\]

Answer 5

I suggest that - at least for the first 50-100 quadratics you will solve - to write down what a,b and c are explicitly because it's very easy to get tricked with them.

Answer 6

in this equation a = 1 b= 1 c = 4

Answer 7

Right, there's a +/- sight instead of multiplication there but I think it's just a typo. Very good though! Right, let's take a look at this equation

Answer 8

oh yea ax^2 + bx +c

Answer 9

Yea it's a typo i meant to put it there =)

Answer 10

You have x^2 - x - 4 =0.

Answer 11

\[x = -1+ - \sqrt{1^2 - 4(4)(1)} \over 2(1)\]

Answer 12

If b were to equal 1 here, how would

x^2 - x - 4

be then different from

x^2 + x - 4

Answer 13

hmmm, u lost me =)

Answer 14

The best way to not get confused by this in the beginning is to write them down like this :

1*x^2 + (-1)*x + (-1)*4 =0. Do you agree that this is the same thing as the original equation you started out with ?

Answer 15

What I am trying to say is that b=-1.

Answer 16

What is wrong with C?

Answer 17

And c=-4.

Answer 18

lol okay hold on lemme write this down and figure it out =)

Answer 19

Take your time!

Answer 20

oh okay i see your point, yea that makes sense =)

Answer 21

Note that b = -1
therefore -b = 1

Answer 22

But if you graph x^2 +x -4 and x^2 - x -4 it is two diff parabolas so i don't see how that could be the same thing =) if u see where i'm coming from =)

Answer 23

@AngusV

Answer 24

My prob is that when i worked out the radicand i got -15 every time and neither of those answers are -15 in the radicand =)

Answer 25

Yes, of course they are not the same thing! What I was trying to say was that they are two different things and they both can't have b=1.

x^2 - x - 4 which has b= -1 is different from
x^2 +x - 4 which has b = 1.

My point was to illustrate that since those two equations aren't equivalent, neither is the b equivalent. And since b is not equivalent, neither are those two. If it makes any sense.

Answer 26

Well, let's look at the radicand thing. There you have the square toot of b^2 - 4*a*c.

Answer 27

so i'm trying to figure out how the sign could possibly change to make it = 17 . . . . . . =)

Answer 28

When you plug in the values you have that the radicand thing equals :

square root of (-1)^2 - 4*1*(-4)

Answer 29

Oh okay i get your non - equality point between those two equations =)

Answer 30

right

Answer 31

Well, you forget that c is -4. That's why it changes the thing to 17. When you solve that you get that the radicand thing equals

square root of 1 - (-16) which is square root of 1 + 16 = square root of 17.

Answer 32

All in all, the answer is C.

Answer 33

Ahhhhh i see, okay thanks so much for the help i think i finally get this =)

Answer 34

I agree that C is a good choice. I used the "completing the square method" which I find easier for simple quadratics than the Quadratic fromula.

Answer 35

In fact the Quadratic Formula was derived using the "completing the square"

Answer 36

lol okay thanks so much, =)