Find another way to solve this problem: (algebra)?

Question

The value of c is the value of a^2 + b^2 when a=2 and b= -2.

Find the value of c^2 - 4.

Find another way to solve this problem.

I did a^2 + b^2 = c
2^2 + (-2)^2=c
4 + 4 + c
c = 8

so c^2 - 4 = v
8^2 - 4 = v
64 - 4 = v
v = 60

What is another way to solve it??

Answer 1

You already figured out this much:
a² + b² = c = 8

Use the difference of squares rule:
c² - 4
= c² - 2²
= (c + 2)(c - 2)
= (8 + 2)(8 - 2)
= 10 x 6
= 60

Answer 2

If your looking for a more easier way I dont think you'll find one easier then what you did. But....

you could plug c=a^2 + b^2 into the second equation:

[(a^2 + b^2)^2]-4

which expands to the equation a^4 + 2(a^2)(b^2) + b^4 -4

and then plug the values for a and b in:

(2)^4+ 2(2^2)(-2^2) + -2^4 -4= 60

Like I said, that way is much more strenuous

Answer 3

Well you could carry the symbols all the way through and substitute later.

c = a^2 + b^2
c^2 = (a^2 + b^2)^2
c^2 = a^4 + 2a^2b^2 + b^4
c^2 - 4 = a^4 + 2a^2b^2 + b^4 = 2^4 + 2*(2^2)*(-2^2) + -2^4 - 4 = 16 + 32 + 16 - 4 = 60.

c^2 - 4 = 60.

Notice that here you never determine c.

Answer 4

c^2-4 is the difference of two squares...

So (c+2)(c-2) is an answer...

c=8

(c+2)(c-2)=(8+2)(8-2)=(10)(6)=60

Answer 5

v = c^2 -- 4
= (a^2 + b^2)^2 -- 4
= {(a + b)^2 -- 2ab}^2 -- 4
= {(2 -- 2)^2 -- 2(2)(--2)}^2 -- 4
= { 0 + 8}^2 -- 4
= 64 -- 4
= 60

Answer 6

"accomplishing the floor" makes the top at that factor 0. Plug in 0 for h and remedy to t (time). h = -5t^2 + 30t 0 = -5t^2 + 30t 0 = -5t(t - 6) Use your 0 product components to set each and each component to 0. -5t = 0 t = 0 (beginning factor) t - 6 = 0 t = 6 6 seconds. desire that facilitates!