Trig homeowrk help!!?

Question

how do you solve...

1) 9x^2+12x^4

2) 36x^2y^6+12x^5y^5+30x^4y^7

3) 1/4 minus square root of 7

Answer 1

1) does 9x^2 + 12x^4  = 0?

take square root of both sides

3x + (square root 12) x^2 = 0

now you can plug it into the quadratic formula!

-3 + or - square root (3^2 - 4 (suare root 12)(0)

all divided by 2(3)

x=2 or x=1

2) 36x^2 y^6 + 12x^5 y^5 + 30x^4 y^7 = 0

take out 6x from the entire equation:

6x (6x y^6  +  2x^3 y^5  +  5x^3 y^7) = 0

take out y^5:

(6xy^5) (6xy + 2x^3 + 5x^3 y^2) = 0

one of the parenthesis has to = 0

(6xy^5) = 0

6x = 0 or y^5 = 0

for this to be true, x or y  = 0

if you plug in x = 0 or y=0 into the original equation, it doesn't matter what the other one equals

so if x = 0 then y = all real numbers

and if y = 0 then x = all real numbers

3) 0.25 - square root of 7

is just calculator work