Question

(6x^2-5)^2

Answer 1

solve for the square of binomial

Answer 2

You have to expand this? Write as (6x²-5)(6x²-5) and do FOIL.

Answer 3

okay cool :)

Answer 4

36x^4-60x^2+25

Answer 5

Yeah right :)

Answer 6

can yo guys help me with one more?

Answer 7

Sure.

Answer 8

it's a radical equation
sqrt (x)+24 - sqrt(x)-15 = 3

Answer 9

No solution

Answer 10

sqrt(x) and - sqrt (x) cancel out.

Answer 11

If it is \[\sqrt{x+24}-\sqrt{x-15}=3\] then it's another story...

Answer 12

Yeah lol :P

Answer 13

yea thats what the equation is

Answer 14

Why did I thought so already? ;)

Ok, now there are square roots involved, so squaring comes to mind...

Answer 15

It's like (a-b)²=3².
So a²-2ab+b²=9
But that means we still haven't got rid of the radicals:\[(\sqrt{x+24})^2-2\sqrt{x+24}\sqrt{x-15}+(\sqrt{x-15})^2=9\]

Answer 16

It would be better if you shifted that stuff on the other side to avoid complication.

Answer 17

But that is not so bad if you look into it:\[x+24+2\sqrt{(x+24)(x-15)}+x-15=9\]

Answer 18

So now I'll do the shifting:\[2x+9=9-2\sqrt{(x+24)(x-15)}\]
The 9's are gone now.
Now square again...

Answer 19

okay Im following along

Answer 20

\[4x^2=4(x+24)(x-15) \Leftrightarrow x^2=x^2+9x-360\]It keeps on getting simpler!

What's next?

Answer 21

is that the answer then? or we have to simplify more?

Answer 22

This is the part where you give the final solution! :)
Subtract x² from both sides and solve for x.

Answer 23

ugh I'm so confused :/

Answer 24

If you subtract x² from both sides of\[x^2=x^2+9x-360\]you get:\[0=9x-360\]Now first write down what 9x must be, then divide by 9 to get x.

Answer 25

40

Answer 26

Well done!.

Just for fun, put x=40 into your original equation to check if it's true...
If you do this, you'll see that you could have got that answer just by guessing!

Answer 27

the problem is asking for a solution set?

Answer 28

That is a set with only one element: {40}