Question

Which two values of x are roots of the polynomial below x^2 - 11x + 13

Answer 1

Because 13 is a prime number, we cannot use the AC method to simplify.

We can instead use the quadratic formula.

$\frac{-b +/- \sqrt{b^2 - 4ac} }{2a}$

11 +/- √121 - 52 / 2

11 +/- √69 / 2

11 +/- 8.3 / 2

(11 + 8.3)/2 = 9.65

(11 - 8.3)/2 = 1.35

The roots of the polynomial are 9.65 and 1.35.

Answer 2

Answer:

$x=\frac{11+\sqrt{69}}{2}$ and $x=\frac{11-\sqrt{69}}{2}$

Step-by-step explanation:

$x^2-11x+13$

13 is a prime number . we cannot factor it because we cannot find two factors whose product is 13 and sum is -11. Apply quadatic formula to find the x values

Given polynomial is in the form of ax^2+bx+c

a= 1, b= -11 and c=13

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Plug in the values in the formula

$x=\frac{11+-\sqrt{(-11)^2-4(1)(13)}}{2(1)}$

$x=\frac{11+-\sqrt{69}}{2(1)}$

$x=\frac{11+\sqrt{69}}{2}$ and $x=\frac{11-\sqrt{69}}{2}$