Question

Quadratic equations:
Solve for x:
X4 - 5X2 + 4 = 0

Answer 1

of course

Answer 2

look at the equation this way

\[(x^2)^2 - 5(x^2) + 4 = 0\]

or there is a method of substitution where

you let

\[u = x^2\]

so you need to solve

\[u^2 - 5u + 4 = 0\]

both equations can be solved by factoring

If you use the equation involving u, when you get the solutions
substitute x^2 back in and then solve for x

hope this helps

Answer 3

why do I have to use x^2?

Answer 4

well the equation you have been given falls into the group called

equations that are reducible to quadratics...

so instead of having

\[ax^2 + bx + c = 0\]

you have

\[a(x^2)^2 + b(x^2) + c = 0\]

or it can be written as

\[au^2 + bu + c = 0\]

when you get it into a quadratic form it can be solved using

factoring, completing the square or general quadratic formula

Answer 5

um I still don't see why x^2 was used couldn't we have used X^4?

Answer 6

well that of it this way using index laws

\[(x^2)^2 = x^{2 \times 2} = x^4\]

Answer 7

and no you can use the generarl quadratic formula on an equation that is a quartic...

you need to get it to a quadratic form....

and there will be 4 solutions...

and the equation is quite simple to factor...

Answer 8

well in my book that is the example of how to solve one of these things and I guess I have t know this way for the test.

Answer 9

can you finish the problem?

Answer 10

this is the factorization

you can have

\[x^4 - 5x^2 + 4 = (x^2 -4)(x^2 -1) \]

or by making a substitution

\[u = x^2\]

the factored from is

\[u^2 - 5u + 4 = (u -4)(u - 1)\]

so in either case after factoring you need to solve

\[x^2 - 4 = 0..... and......x^2 - 1 = 0\]

Answer 11

thats how you do it.... and you'll find there are 4 solutions.

Answer 12

alright thanks!