solve.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Question

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

Answer 1

Given quadratic equation is 25x^4 - 15x^2 + 2 = 0

To solve this equation let us assume a=x^2, then the equation becomes:

                                                 25a^2 - 15a +2 = 0

                                            => 25a^2 - 5a - 10a + 2 = 0 

                                            => 5a(5a - 1) -2(5a -1) = 0

This factors to                     => (5a - 2)(5a - 1) = 0

                                           => 5a - 2 = 0 and 5a - 1 = 0

                                           => 5a = 2 and 5a = 1

                                           => a = 2/5 and a = 1/5

But a=x^2,

                              => x^2 =a

                              => x^2 = 2/5 and x^2 = 1/5

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

So these are the four roots of the given quadratic equation.

Comments

The solutions are  ± √(1/5) and ± √(2/5).

Answer 2

25^x4 - 15x^2 + 2 = 0

Let x^2 = t

25t^2 - 15t + 2 = 0

25t^2 - 10t - 5t + 2 = 0

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

Take out common factors.

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

5t - 1 = 0 or 5t - 2 =0

5t = 1 or 5t = 2

t = 1/5 or t = 2/5

Substitute t = x^2 in the equation

x^2 = 1/5 or  x^2 = 2/5

x = ± √1/5 or  x = ± √2/5

x = {± √1/5 , ± √2/5}

The solution's are  x = {± √1/5 , ± √2/5}