Question

solve. 25x^4-15x^2+2=0

Answer 1

you have to find a multiple of 25 and a multiple of 2 that when you add gives you -15
then you have 5.5 and 2.1 (5x2=10 and 5x1 = 5) (10+5=15)
then you have
(5x -2) (5x-1)

Answer 2

then you have to equal that to 0
you have
5x-2 = 0 and 5x-1=0 and solve for x

Answer 3

ok after that what i need to do then.

Answer 4

then 5x-2 =0
5x = 2
x = 2/5

the same with 5x-1=0
5x = 1
x = 1/5

Answer 5

25x^4-15x^2+2=0
First do a substitution. Let x^2 = u; then x^4 = u^2.
Replace x with u to get a quadratic equation:
25u^2 - 15u + 2 = 0
This equation ios of the form ax^2 + bx + c = 0
Now multiply ac together: ac = 25*2 = 50
Find two factors of 50 that add up to b, -15:
They are -10 and -5
Break up the middle coefficient into these tow factors and rewrite the equation:
25u^2 - 10u - 5u + 2 = 0
Now take a common factor out of the first two terms, and take out a common factor out of the last two terms:
5u(5u - 2) -1(5u - 2) = 0
Factor out the common term, 5u - 2 to get:
(5u - 2)(5u - 1) = 0
Now substitute x back in (remember, u = x^2, so we get:
(5x^2 - 2)(5x^2 - 1) = 0
Now set each binomial equal to zero:
5x^2 - 2 = 0 or 5x^2 - 1 = 0
5x^2 = 2 or 5x^2 = 1
x^2 = 2/5 or x^2 = 1/5
x = +/-sqrt(2/5) or x = +/- sqrt(1/5)

Answer 6

ok so the answer who will look like..

Answer 7

I mean how