Question

2X^4+X^3-11X^2+X+2=0​

Answer 1

Helo mate here your ans

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((((2•(x4))-(x3))-11x2)-x)+2 = 0

Step 2 :

Equation at the end of step 2 :

(((2x4 - x3) - 11x2) - x) + 2 = 0

Step 3 :

Polynomial Roots Calculator :

3.1 Find roots (zeroes) of : F(x) = 2x4-x3-11x2-x+2

Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 2 and the Trailing Constant is 2.

The factor(s) are:

of the Leading Coefficient : 1,2

of the Trailing Constant : 1 ,2

hope this ans help you..

Answer 2

Answer:

The roots are $2,\frac{1}{2},\frac{-3\±\sqrt5}{2}.$

Step-by-step explanation:

Given :-

$2x^4+x^3-11x^2+x+2=0$

To find :-

Roots of the given equation.

Solution :-

$2x^4+x^3-11x^2+x+2=0$

Divide on both sides by $x^2$.

$\frac{2x^2}{x^2}+\frac{x^3}{x^2}-\frac{11x^2}{x^2}+\frac{x}{x^2}+\frac{2}{x^2}=0$

$2x^2+x-11+\frac{1}{x}+\frac{2}{x^2}=0$

$2[x^2+\frac{1}{x^2}]+(x+\frac{1}{x})-11=0$

$2[(x+\frac{1}{x}^2-2]+(x+\frac{1}{x})-11=0$

Let $x+\frac{1}{x}=a$

$2[a^2-a]+a-11=0$

$2a^2-4+a-11=0$

$2a^2+a-15=0$

$2a^2+6a-5a-15=0$

$2a(a+3)-5(a+3)=0$

$(a+3) (2a-5)=0$

$(a+3)=0\;or\;2a-5=0$

$a=-3\;or\;a=\frac{5}{2}$

Case (i) a = $-3$

$x+\frac{1}{x}=-3$

$\frac{x^2+1}{x}=-3$

$x^2+3x+1=-3$

$x=\frac{-3\±\sqrt9-4}{2}$

∴ $x=\frac{-3+\sqrt5}{2}$

Case (ii) a = $\frac{5}{2}$

$x+\frac{1}{x}=\frac{5}{2}$

$2x^2+2=5x$

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

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

∴ $x=2\;or\;\frac{1}{2}$

∴ The roots are $2,\frac{1}{2},\frac{-3\±\sqrt5}{2}.$