Answer-
The values of x, which are the roots of the polynomial are
[tex]\dfrac{11+ \sqrt{69}}{2},\ \dfrac{11- \sqrt{69}}{2}[/tex]
Solution-
The given polynomial,
[tex]=x^2 - 11x + 13[/tex]
We know that,
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here,
a = 1, b = -11, c = 13
Putting the values,
[tex]x=\dfrac{-(-11)\pm \sqrt{(-11)^2-4\times 1\times 13}}{2\times 1}[/tex]
[tex]=\dfrac{11\pm \sqrt{121-52}}{2}[/tex]
[tex]=\dfrac{11\pm \sqrt{69}}{2}[/tex]
[tex]\therefore x=\dfrac{11+ \sqrt{69}}{2},\ \dfrac{11- \sqrt{69}}{2}[/tex]
