Can someone help me solve 25x^4-15x^2+2=0? Thank you!?

Question

Can someone help me solve 25x^4-15x^2+2=0? Thank you!?

Answer 1

use the quadratic formula. x= (-b+_ sqrt b^2 + 4ac)/2a

b= -15

a=25

c=2

Answer 2

25 x^4-5 x^2-10 x^2+2=0

or 5x^2(5x^2-1)-2(5x^2-1)=0

or (5x^2-2)(5x^2-1)=0

or 5x^2=2 or 5x^2=1

or x^2=2/5 or x^2=1/5

or x=+-sq rt(2/5) or x=+-sq rt(1/5) ANS

Answer 3

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

Quadratic Equation: Ax + By + C = 0

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

Ax + By + C = 0

Now look and see that your letters match up with certain numbers....it will always be like this!

A = 25, B = -15, C = 2

Now insert these into the Quadratic Formula

x = -b ± the square root of b^2 - 4ac all divided by 2a

x = -(-15) ± the square root of (-15)^2 - 4(25)(2) / 2(25)

x = 15 ± the square root of 125 - 200 / 50

x = 15 ± the square root of -75 / 50

x = 15 ± 5i times the square root of 3 / 50

x = 15 / 50 ±  5i√3 /50

x = 3 / 10 ±  5i√3 /50

Your answer is: x = 3 / 10 ±  5i√3 / 50

Hope this was helpful

Answer 4

believe it or not, you can use the quadratic equation to solve this

set u=x^2, then your equation becomes:

25u^2-15u +2=0

this factors to

(5u-1)(5u-2)=0

or u=1/5 and u=2/5

but u=x^2, so we have that

x=+/-sqrt[1/5] and x=+/-sqrt[2/5

so these are the four roots of your original equation

Answer 5

divide everything by 25:

x^4 - (3/5)x^2 + 2/25 = 0

factors of 2/25 include (-2/5) and (-1/5), and these add to give -(3/5)

so the eqn then is,

(x^2 - 2/5)(x^2- 1/5)

slns are when

x^2 = 0.4

and when,

x^2 = 0.2

therefore the slns are four slns which are (-sqt(0.4)), (sqrt(0.4)), (-sqrt(0.2), and sqrt(0.2)  

Answer 6

=(5x^2 -2)(5x^2 -1) =0

5x^2 =2, 5x^2 =1

x^2 =2/5, x^2 =1/5

x=+/- (sqrt 10)/5, +/- (sqrt 5) /5