Algebraic problem

If ab = -5 and (a +b)^2 = 16, then a^2 + b^2 =

can i plug in first ?

Psychguy98 said:

If ab = -5 and (a +b)^2 = 16, then a^2 + b^2 =

can i plug in first ?

Click to expand...

Toasters and lamps get "plugged in."

Here's how you might approach this problem.

You know that

(a + b)[sup:3m8ngawx]2[/sup:3m8ngawx] = 16

But (a + b)[sup:3m8ngawx]2[/sup:3m8ngawx] means (a + b)(a + b). What do you get if you DO that multiplication?

Try it.

Do you see some of the terms you either know something about, or are asked to find?

I really can't offer any other hints without handing you the answer...

does that equal 2ab?

from (a +b) (a + b) or can (a +b)^2 = a^2 +2ab + b^2?

Psychguy98 said:

from (a +b) (a + b) or can (a +b)^2 = a^2 +2ab + b^2?

Click to expand...

Ok, you have done the multiplication correctly.

(a + b)[supgyxh57y]2[/supgyxh57y] = a[supgyxh57y]2[/supgyxh57y] + 2ab + b[supgyxh57y]2[/supgyxh57y]

From the given, you know that (a + b)[supgyxh57y]2[/supgyxh57y] = 16, and that ab = -5. SUBSTITUTE 16 for (a + b)[supgyxh57y]2[/supgyxh57y]. SUBSTITUTE -5 for ab:

(a + b)[supgyxh57y]2[/supgyxh57y] = a[supgyxh57y]2[/supgyxh57y] + 2 ab + b[supgyxh57y]2[/supgyxh57y]

16 = a[supgyxh57y]2[/supgyxh57y] + 2(-5) + b[supgyxh57y]2[/supgyxh57y]

Now....can you find the value of a[supgyxh57y]2[/supgyxh57y] + b[supgyxh57y]2[/supgyxh57y] from that? (Hint: get a[supgyxh57y]2[/supgyxh57y] + b[supgyxh57y]2[/supgyxh57y] on one side of the equation....)

(a + b)^2 = a^2 + 2 ab + b^2

16 = a^2 + 2(-5) + b^2

16 = a^2 -10 + b^2
+10 +10

26 = a^2 + b^2