SOLUTION: 25x^4-25x^2+6=0 i worked the problem out and got x=square root of 3/5 and 2/5 am i doing this right

Question 443190: 25x^4-25x^2+6=0
i worked the problem out and got x=square root of 3/5 and 2/5 am i doing this right
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.6, 0.4. Here's your graph:

It is not the square root of 2/5 and 3/5, but simply 2/5 and 3/5.
RELATED QUESTIONS
4over square root 5 +square root 3 I worked it to 4(square root 5 -square root 3) over 2 (answered by jim_thompson5910)

i need help on this problem. i am supposed to solve the equation and check for extraneous (answered by Alan3354,jim_thompson5910)

What is the factor of 5x^2 - 25x=0? Is it 5x(x -5)? Am i... (answered by stanbon)

I need help factoring out this problem{{{20x^3+16x^2+25x+20=0}}} I did it in two group... (answered by jim_thompson5910,Alan3354)

Problem: 5X^2+75=25X What I've got so far: 5X^2-25X+75=0 (5x- )(x- ) or... (answered by scott8148)

25x^4-25^2+6=0 The solution is x= The answer that I got is: sqrt15/5,-sqrt15/5 (answered by edjones)

Use the rational root theorem and the factor theorem to help solve the following... (answered by stanbon)

I am having problems solving fractional equations.This is the problem i'm working... (answered by ewatrrr)

Here are a few problems that I am not sure if I got them right. Thanks. Paul 1)... (answered by ankor@dixie-net.com)