Help w/ algebra 3/4?!?

Question

how do you solve this equation: xSquared + x - 12 = 0

i have a couple equations in that form, and i just need help with that one to get me started. thank you:]

Answer 1

x^2 + x - 12 = 0

You factor.

(x-3)(x+4) = 0

x = 3, -4

Check answers.

3^2 + 3 - 12 = 0

9+3 - 12 = 0

True

-4^2 - 4 - 12 = 0

16 - 4 - 12 = 0

True

Answer 2

Pranil  answered this 3 months ago and got best answer

in the mathematics section

Best Answer - Chosen by Voters

x² + x – 12

find the factors of 12 such thet their sum or difference is 1

3 × 4 = 12 and 4 – 3 = 1

x² + x – 12

= x² + 4x – 3x – 12

= x(x + 4) – 3(x + 4)

= (x – 3)(x + 4)

x = 3 or x = – 4

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Answer 3

It is a quadratic equation, you need to factor it and get it in the form

(x-A)(x-B)=0, so the answer will then be, x=A and x=B

For that example

(x+4)(x-3)=0, so x=-4 and x=-3 is your answer.

If you write your equation in this form:

Dx(squared) + Ex + F =0

then you need (x-A)(x-B)=0,

where A+B=E and AB=F

use that trick and you should always be able to factor the formula and get the answer..

The trick is that, A+B

Answer 4

I would factor it (or you can use quadratic formula).

(x + 4) (x - 3) = 0 .....I found factors of 12 that had a difference of 1

Set each factor equal to zero (zero product property)

x + 4 = 0 or x - 3 = 0

x = -4 or x = 3

Hope this helps !

Answer 5

factor the top one

x^2 + x - 12 = 0

(x + 4) (x - 3) = 0

which means x = -4, x = 3

2sqrt(32) = 8sqrt(2)

6sqrt (8)  = 12 sqrt (2)

sqrt 128  = 8 sqrt (2)

all of these have perfect squares that can be taken out from the radical

so it would be 8sqrt(2) + 12 sqrt(2) - 8sqrt(2)

= 12 sqrt (2)

Answer 6

(x+4)(x-3)

x=-4   x=3

8radical2+12radical2-8radical2=

12radical2