Question

@ganeshie8 can you help me? i'll post the question below, it's a quadratic formula question...

Answer 1

\[3x ^{2}+7x-24+13x\]

Answer 2

Use the quadratic formula to solve and use factoring to check your answer (I already know the answer is -2,4 but i keep getting 1,3

Answer 3

that should be = 13x not +13x

Answer 4

@ilikequadraticsokay Please close this thread given that it is the same question which you posted. Thanks.

http://openstudy.com/study#/updates/529e87dce4b0ad25f2dee29a

Answer 5

3x^2 + 7x - 24 = 13x
3x^2 + 7x - 13x - 24 = 0
3x^2 - 6x - 24 = 0
divide by 3
x^2 - 2x - 8 = 0
discriminant = b^2 - 4ac = (-2)^2 - (4)(1)(-8) = 4 + 32 = 36
roots = { -(-2) +/- sqrt(36) } / 2 = (2 +/- 6) / 2 = 1 +/- 3
= -2 or 4

Answer 6

i dont know what the discriminant is

Answer 7

x^2 - 2x - 8 = 0
factoring
x^2 - 4x + 2x - 8 = 0
x(x - 4) + 2(x - 4) = 0
(x - 4)(x + 2) = 0
x = -2 or +4

Answer 8

just the term that goes inside the square root in the quadratic formula: b^2 - 4ac is called the discriminant.

Answer 9

Oh okay, one moment.

Answer 10

How did you go from 1 +/- 3 to -2,4?

Answer 11

1 + 3 = 4
1 - 3 = -2

Answer 12

That makes sense.

Answer 13

alright.

Answer 14

So I was just forgetting to actually solve the +/- part?

Answer 15

not sure. if you retrace your steps you may be able to spot the mistake.

Answer 16

x^2 - 2x - 8 = 0
\[\Large \frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a } = \frac{ -(-2) \pm \sqrt{(-2) ^{2} - 4(1)(-8)} }{ 2(1) }\]

Answer 17

\[= \frac{ 2 \pm \sqrt{4 + 32} }{ 2 } = \frac{ 2 \pm \sqrt{36} }{ 2 } = \frac{ 2 \pm 6}{ 2 } = 1 \pm 3\]

Answer 18

1 + 3 = 4
1 - 3 = -2

Answer 19

Yeah, I did all that. I guess I just thought it ended at 1 +/- 3

Answer 20

oh, ok. you will find two roots: one by adding and another by subtracting the second number.

Answer 21

Okay, thank you so much!

Answer 22

you are welcome.