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Kipish [7]
3 years ago
8

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

Mathematics
2 answers:
matrenka [14]3 years ago
7 0

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

<h3>The roots of the polynomial are 9.65 and 1.35.</h3>
Ad libitum [116K]3 years ago
3 0

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}

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