Question

x^2-5x-14

Answer 1

This is a quadratic equation. So we need to use the quadratic formula. Do you know how to use it?

Answer 2

no

Answer 3

@jcpd910

Answer 4

Okay, I'll walk you through it.

Answer 5

\[x = \frac{ -b \sqrt{b^{2}-4ac}}{ 2a }\]

Answer 6

That's the quadratic formula. Looks daunting right?

Answer 7

can u show ur steps on how to solve? @jcpd910

Answer 8

Of course, it's not that hard. Here's your equation:
ax^2 + bx + c = 0

x^2 - 5x - 14

So tell me what a, b, and c are. It's okay if you can't, I'll show you how if you don't know.

Answer 9

i honestly don't know @jcpd910

Answer 10

Okay, that's fine. Can you tell me what number x^2 has in front of it?

Answer 11

?

Answer 12

x^2 = 1x^2

Answer 13

ohhh

Answer 14

It's confusing but then you get it ik :)

Answer 15

ax^2 + bx + c = 0
1x^2 - 5x - 14 = 0
Do you see it now?

Answer 16

yes

Answer 17

what do i do next

Answer 18

tell me the values of a, b, and c.

Answer 19

a=1,b=5,c=14

Answer 20

Almost, but you see how it says - 5 and - 14?

Answer 21

oh okay!

Answer 22

Subtracting is just adding a negative number: 1 - 1 = 1 + (-1)

Answer 23

So what are the values?

Answer 24

what do i do? subtract 1-5-14?

Answer 25

No, haha, I wish it were that easy. One second bbrb

Answer 26

I'm back

Answer 27

Now that you have the values, you plug them into the equation. emc can I take this one please?

Answer 28

@jcpd910 there is no = so it is not an equation. Now, if the instructions say to factor it, you can assume it is one that is set to 0.

Answer 29

Oops, e.m is right. Sorry about that lub. How do we solve it then e.m?

Answer 30

Well, does it need to be factored or what? It is just sort of there.

Answer 31

That's why I assumed it was = 0 for a quadratic, I didn't think to just factor it out though.

Answer 32

it needs to be factored

Answer 33

Yah. Depends on if you need roots, need to factor, or what. Those are closely related things, but the answers are slightly different.

Answer 34

and on a totally unrelated note, how do u become a moderator?

Answer 35

\(Ax^2+By+C\) factored to some sort of \((x\pm m)(x\pm n)\)
When A = 1, which yours does, there is a little logic:
If C is positive, both m and n have the same sign.
If C is negative, m and n have different signs.
If B is positive, the larger or both of m and n must be positive.
If B is negative, the larger or both of m and n must be negative.

You can google up becoming an openstudy moderators. We have talked about it lots of times in feedback.

Answer 36

OK, so, lets look at yours: \(x^2-5x-14\)

Both B and C are negative. So these two rules:
If C is negative, m and n have different signs.
If B is negative, the larger or both of m and n must be negative.

Now for the other little thing you want to find.

Answer 37

You want two numbers, I called them m and n, that multiply to become C and add to become B.

So \(mn=-14\) and \(m+n=-5\)

That means you can factor 14 different ways to get possible answers.

Answer 38

I assume that to become a moderator you have to be helpful, positive, and don't ask for moderator. That's normally how it works. :)

Answer 39

\(1\times 14=14\) If the larger must be negative, that would be \(1\times -14=-14\) BUT: \(1+(-14)=-13\) so those are not your factors. They fit one result but not the other.

Answer 40

omg i get it i get it! i know the answer!!

Answer 41

thanks so much for y'all help!

Answer 42

Wait, the quadratic formula wouldn't have worked but we can still factor?

Answer 43

Hehe. OK, I think you see it now. What do you think it is?

Answer 44

can i give medals to both or just one person

Answer 45

Give it to me, I gave one to e.m

Answer 46

The quadratic would. It is just that you need to flip the sign when you plug them into the factors.

Answer 47

(x+2)(x-7)

Answer 48

Yes. That is the answer.

Answer 49

bingo.

Answer 50

Here is the quadratic method for reference.

\(x = \dfrac{ -b \pm\sqrt{b^{2}-4ac}}{ 2a }\)

\(x = \dfrac{ -(-5)\pm \sqrt{(-5)^{2}-4(1)(-14)}}{ 2(1) }\)

\(x = \dfrac{ 5 \pm \sqrt{25+56}}{ 2 }\)

\(x = \dfrac{ 5 \pm \sqrt{81}}{ 2 }\)

\(x = \dfrac{ 5 \pm 9}{ 2 }\)
so
\(x = \dfrac{ 5 + 9}{ 2 }\) and \(x = \dfrac{ 5 - 9}{ 2 }\)

\(x = \dfrac{ 14}{ 2 }\) and \(x = \dfrac{ -4}{ 2 }\)

\(x = 7\) and \(x = -2\)

But in factored form you move them over to multiply:

\(x - 7=0\) and \(x + 2=0\)

\((x - 7)(x + 2)\)

Answer 51

Now you can see why a little logic for factoring can save you a lot of work over using that all the time!