Disguised Quadratic equations?

Question

a.) By using substitution p = x + (1/x), show that the equation 2x^4 + x^3 - 6x^2 + x + 2 = 0 reduces to 2p^2 + p - 10 = 0

b.) Hence solve 2x^4 + x^3 - 6x^2 + x + 2 = 0.

Answer 1

2p² + p - 10 = 0
(2p² - 4p) + (5p - 10) = 0
2p(p - 2) + 5(p - 2) = 0
(2p + 5)(p - 2) = 0
p = -5 / 2 or p = 2.

If p = x + 1 / x, then
2p² + p - 10 = 0 implies
2(x + 1 / x)² + (x + 1 / x) - 10 = 0,
2(x² + 2 + 1 / x²) + (x + 1 / x) - 10 = 0,
2x² + 4 + 2 / x² + x + 1 / x - 10 = 0,
2x² + x - 6 + 1 / x + 2 / x² = 0.
Multiplying through by x²,
2x^4 + x³ - 6x² + x + 2 = 0.

Therefore, 2x^4 + x³ - 6x² + x + 2 = 0 has solutions
p = -5 / 2 or p = 2, where p = x + 1 / x.

x + 1 / x = -5 / 2 implies
2x² + 2 = -5x
2x² + 5x + 2 = 0
(2x² + 4x) + (x + 2) = 0
2x(x + 2) + 1(x + 2) = 0
(2x + 1)(x + 2) = 0
x = -1 / 2 or x = -2.


x + 1 / x = 2 implies
x² + 1 = 2x
x² - 2x + 1 = 0
(x - 1)² = 0
x - 1 = 0
x = 1.

Therefore, x = -2 or x = -1 / 2 or x = 1.

Answer 2

2p^2+p-10=0 =>2[(x^2+1)^2/x2]+(x^2+1)/x-10=0 =>
2x^4+x^3-6x^2+x+2=0 hence proved
2p^2+p-10=(2p+5)(p-2)=0
so p=-5/2 or 2
x^2+1=-5/2x =>2x^2+5x+2=0
(2x+1) (x+2)=0
so the roots are 1,-1,-2 and 1/2

Answer 3

OK, what is your question? Are you asking if you can substitute like you did to make the solution easier to find? Sure you can, what you are doing is changing the coordinate system. Scientists, engineers, and mathematicians do it all the time to solve complex equations.

There are even some commonly used operators, called transforms, that do some really neat coordinate changes. Z-transforms, Laplacians, and others belong in this category. In fact there is a whole branch of math, called curvilinear coordinates, that deals with coordinate transformation.

By the way, I didn't check your algebra (your coordinate transformation); but I'm assuming you did it right. But recognize, once you have a value (values) for p, you need to use your transformation to solve for x if an answer in x is what you want. Solving back for x, using p, is called an anti-transform. Anti-Laplacians, for example, return a solution back to the original coordinate system.

Answer 4

Dividing of the first equatin by x^2 gives :
2x^2+x-6+1/x +2/x^2 = 0

from p=x+1/x => p^2=x^2 +2*x*(1/x)+1/x^2
=> p^2 =x^2+1/x +2

so ,from the first equation :

2x^2+x-6+1/x+2/x^2=0
=> 2x^2+4+2/x^2 - 10 +x +1/x=0
=> 2*(x^2+2+1/x^2) +(x+1/x) -10 =0
=> 2*p^2+p-10=0

solving for p:

p1=(-1+sqrt(1+4*2*10)) /4
p2=(-1-sqrt(1+4*2*10)) /4

=> p1= (-1+9)/4=2
p2=(-1-9)/4=-2.5

now solving for x:

p=x+1/x => p*x=x^2+1 = > x^2-px+1=0

x1=( p+sqrt(p^2-4))/2
x2=( p -sqrt(p^2-4)/2

for p=2 :
x1=x2=(2+sqrt(4-4))/2=1

for p=-2.5:
x1=(-2.5 + sqrt(6.25-4))/2 = (-2.5 + sqrt(2.25))/2 = (-2.5+1.5)/2 =- 0.5
x2=(-2.5-1.5)/2= -2.

So , at the end there are 3 solutions :
x1=-1
x2=-2
x3=-0.5

Answer 5

Multiply through by x^1/2 to give x + 10 = 7x^1/2 let t = x^1/2 t^2 + 10 = 7t rearrange t^2 - 7t + 10 = 0 factorize (t - 5)(t - 2) = 0 Either t - 5 = 0 => t = 5 => x^1/2 = 5 => x = ? OR t - 2 = 0 => t = 2 => x^1/2 = 2 => x = ?

Answer 6

Remember, your equation to the 4th power has 4 solutions... either real or imaginary.