CPhill

I saw that after I had posted...let me give Aziz full credit too!!   (I won't take any points for my repeated answer!!.......)

I agree with Melody's answer......however......I'll bet that 99% of us would have worked it just like ND did.......(I know I would have !!!)

Let me add one thing to what Melody is saying......we can always "test" whether a function is even by replacing x with -x.....if the reults are the same, then the function is even.

Then, is 2^(x) = 2^(-x)  ???........NO !!!

Note that a function like x² is even, because...

x² = (-x)²  

And this is exactly what Melody said......

I'm assuming that you might be trying to find the missing side of a right triangle with a leg of 4cm and a hypoteneuse of 64cm.....so, squaring both sides, we have......

4096 = x^2 + 16   subract 16 from both sides

4080 = x^2      take the positive root of both sides

√4080 = x = √(16*255) = 4√255 cm..... or about 63.87 cm

This appears to be...

1¹,  2² , 3³ , 4⁴ , 5⁵ 

So the missing value is 4⁴ = 256

It's a shorthand way for expressing a number without using scientific notation directly....in this case.....

2.0574749e+21  equates to.....

2.0574749 x  10²¹   

P.S.    ......This is always confusing to me, too!!!

OK......so we have

2x^4-2x^2-12 = 0   divide everything by 2

x^4 - x^2 - 6 = 0    we can factor this directly as

(x^2 + 2 ) (x^2 - 3) = 0

Setting the first two factor to 0, we have that x^2 + 2 = 0  ... subtract 2 from both sides....

x^2 = -2      take the square root of both sides

x = ±√(-2) = ±i√(2) or ±√(2)i, if you prefer  .....and those are the complex roots.....

And setting  x^2 - 3 = 0 and adding 3 to both sides, we get

x^2 = 3     take the square root of both sides and we get......

x = ±√(3)  ....and those are the real roots....

What's the question???

√2 *√15 = √30

We might also use a trig identity to solve this....note that sin(pi/6)*cosx-cos(pi/6)*sinx= sin(pi/6 -x)......so we have

sin(pi/6 - x) = -1/2

Since the sinθ = -1/2 when θ = 7/6pi, we can solve (pi/6 - x) = 7/6pi, and find that x = -pi. But, -pi is the same coterminal angle as pi. So x = pi is one answer (as Aziz found). As a check, note that sin(pi/6 - pi) = sin (-5/6pi) = -1/2

Also, sinθ = -1/2 when θ = 11/6pi. And in a like manner, we have (pi/6 - x) = 11/6pi, and we find that x = -10/6pi or -5/3pi. But -5/3pi is the same coterminal angle as pi/3... (again, as Aziz found). As a check, note that sin(pi/6 - pi/3)  = -1/2.

Using the identity saves a bit of work....but you also obtain answers that you must "convert" to stay within the specified interval....!!! Your choice....