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 $\frac{-1-i\sqrt{3}}{2}$, $\frac{-1+i\sqrt{3}}{2}$