Answer to Question #177126 in Math for Mary

Question

 Given the quantic equation below, solve it to find the values of x. (3pts) 

 

 2𝑥5 − 6𝑥3 − 4𝑥2 - 2𝑥 + 4 = 0 

Answer 1

Divide both sides of the equation by the common factor 2:

$x^5-3x^3-2x^2-x+2=0$

Use Rational Root Test to find rational roots. The factors of the constant term (2) are 1, 2, -1,-2. Check whether these numbers are roots of the equation:

$1^5-3\cdot1^3-2\cdot 1^2-1+2=1-3-2-1+2=-3\mathrlap{\,/}{=}0$

x=1 is not a solution.

$2^5-3\cdot2^3-2\cdot 2^2-2+2=32-24-8-2+2=0$

x=2 is a solution and x-2 is a factor of the equation. Factorize it:

$x^5-3x^3-2x^2-x+2=$ $=x^5-2x^4+2x^4-4x^3+x^3-2x^2-x+2=$ $=x^4(x-2)+2x^3(x-2)+x^2(x-2)-1(x-2)=$ $=(x-2)(x^4+2x^3+x^2-1)=$ $=(x-2)((x^2+x)^2-1)=$ $=(x-2)(x^2+x-1)(x^2+x+1)$

Now the equation has the form:

$(x-2)(x^2+x-1)(x^2+x+1)=0$

1) Solve the equation $x^2+x-1=0$

$x_1=\frac{-1-\sqrt{1-4\cdot1\cdot(-1)}}{2}=\frac{-1-\sqrt{5}}{2}$ $x_2=\frac{-1+\sqrt{1-4\cdot1\cdot(-1)}}{2}=\frac{-1+\sqrt{5}}{2}$

2) Solve the equation $x^2+x+1=0$

The Discriminant is negative: $D=1-4\cdot1\cdot1=1-4=-3$, so the equation has no real solutions, but has complex solutions:

$x_1=\frac{-1-\sqrt{1-4\cdot1\cdot1}}{2}=\frac{-1-\sqrt{-3}}{2}=\frac{-1-i\sqrt{3}}{2}$ $x_2=\frac{-1+\sqrt{1-4\cdot1\cdot1}}{2}=\frac{-1+\sqrt{-3}}{2}=\frac{-1+i\sqrt{3}}{2}$

Answer: The equation has 3 real solutions: 2, $\frac{-1-\sqrt{5}}{2}$, $\frac{-1+\sqrt{5}}{2}$ and 2 complex solutions $\frac{-1-i\sqrt{3}}{2}$, $\frac{-1+i\sqrt{3}}{2}$