To calculate: The solution set of the polynomial equation 11v−2+23v−1+2=0

Question

To calculate: The solution set of the polynomial equation  11v−2+23v−1+2=0

Lauren Fuller · Accepted Answer

Calculation:
This is a polynomial equation with one side equal to zero.

l l y^{- 2} + 23 v^{- 1} + 2 = 0 S

Let v-! = 4, the equation becomes,

l l y^{2} + 23 v^{1} + 2 = 0 S

Simplify the given equation,

11 t^{2} + 23 t + 2 = 0

11 t^{2} + 22 t + 1 t + 2 = 0

11 t (t + 2) + 1 (t + 2) = 0

(t + 2) (11 t + 1) = 0

Set each factor equal to zero.

(t + 2) = 0

t = - 2

Or

(11 t + 1) = 0

t = \frac{- 1}{11}

Now, the value of vis,

v = t^{- 1}

v = - \frac{1}{2} r=-1 r = - 11

Check at

v = \frac{1}{2}

11 {(- \frac{1}{2})}^{- 2} + 23 {(- \frac{1}{2})}^{1} + 20

11 {(- 2)}^{2} + 23 {(- 2)}^{1} + 20

11 (4) - 46 + 20

0 = 0

The left side value is equal to the right-side value. Thus, the value

v = - 11

is verified.
Therefore, all together the solution set of polynomial equation

11 v^{- 2} + 23 v^{- 1} + 2 = 0

is

{- \frac{1}{2}, - 11}

To calculate: The solution set of the polynomial equation 11v^{