Question

Solve: b^2+b-12=0.

I came up with -4,3

Answer 1

you can substitute the values you got back into the equation and see if that works

Answer 2

(-4)^2 +(3)-12=0
16+3-12=0
19-12=17
Is this correct?

Answer 3

no you would have to substitute two different times - -4 and 3 both are answers, but the equation has two different solutions

Answer 4

Checking b = -4

b^2+b-12=0

(-4)^2+(-4)-12=0

16 - 4 - 12 = 0

12 - 12 = 0

0 = 0 ... so the value b = -4 is a solution

Answer 5

so the right way of verifying your solution would be:
(-4)^2 -4 -12 = 0 check this

and do the same for 3

Answer 6

Do the same with the other value to confirm

Answer 7

3^2+3-12=0
9+3-12=0
12-12=0
Is this right?

Answer 8

good, so you've confirmed that b = 3 is also a solution

Answer 9

So the answer is -4,3 right?
Or do I say b=-4, b=3

Answer 10

you would say the solutions are b = -4 or b = 3

Answer 11

Some books use the idea of the solution set, which means you'd write {-4,3}. But the first method works just fine.

Answer 12

I think my book goes by solution set...You guys are awesome;-) Thanks!!!

Answer 13

you're welcome