b^2 - 49 =0....WHAT?!?!?

Question

AND...

X^2 -9X=0

B^2 - 4B = 32

AND IT SAYS THEIR SHOULD BE "TWO" ANSWERS! HOW IN THE WORLD DO I DO THIS!

Answer 1

b^2=49
b= +or- 7

b^2-4b-32
(b-8) (b+4)
b=8 and -4

Answer 2

> b^2 - 49 =0....WHAT?!?!?

Take a deep breath, relax :-)
It just means: square some number, then subtract 49, and the result is zero. If you think about it, that must mean that b² is 49. (Because then you have: 49–49, which indeed is zero.

So, what number do you have to square to get 49?

> THEIR SHOULD BE "TWO" ANSWERS!

Yes. There are actually two DIFFERENT numbers which you can square to get 49. (Hint: one of them is negative!)

> X^2 -9X=0

That can be rewritten as: x² = 9x

Now, you can simplify that by dividing both sides by x (BUT only if x is not zero, since you are not allowed to divide by zero). So:

If x is not 0:
x²/x = 9x/x
x = 9

So there's ONE answer. Now let's see what happens if x IS zero:

x² = 9x
0² = (9)(0)
0 = 0

That works too. So the two answers to that one are: x=9 and x=0

> B^2 - 4B = 32

This one's a little trickier. First, subtract 32 from both sides, and you get:
B² – 4B – 32 = 0

Now, try to factor the left side. You should end up with something like this:
(B – [something])(B – [something else]) = 0

(I'm not going to teach you how to factor--that's something you should know by now from your class.)

Once you get it factored, notice that if you set B either to the [something] or the [something else], the equation comes out to zero. So those are your two solutions.

Answer 3

b^2 - 49=0
b^2=49
take the square root of both sides
b=+ or -7

x^2 - 9x=0
x(x-9)=0
x=0 and x-9=0 so...
x=0 and 9

b^2 - 4b=32
b^2-4b-32=0
(b-8)(b+4)=0
b=8 and -4

Answer 4

b^2 - 49 =0
b^2 = 49
b = -7 or b = +7

X^2 -9X=0
X(X - 9) = 0
X = 0 or X = 9

B^2 - 4B = 32
B = 8 or B = -4