The roots of f(x) are 7+root 30 and 7 -root 30.
We have given that the quadratic equation is
[tex]f(x)=x^2-14x+19[/tex]
We have to find the zeros of the given f(x)
Compaire the given equation with the general form of quadratic equation
What is the general form of quadratic equation?
[tex]ax^2+bx+c=0[/tex]
So we get,a=1,b=-14,c=19
We find the roots by using the quadratic formula method
What is the quadratic formula ?
[tex]x_{1,2}=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex]
Use the value of a,b and c in above formula we get,
[tex]x_{1,2}=\frac{-(-14)\±\sqrt{(-14)^2-4(1)(19)} }{2(1)}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{30}}{2\cdot \:1}[/tex]
[tex]x_1=\frac{-\left(-14\right)+2\sqrt{30}}{2\cdot \:1}[/tex]
[tex]=\frac{14+2\sqrt{30}}{2}[/tex]
[tex]=\frac{14+2\sqrt{30}}{2}[/tex]
[tex]=\frac{2\left(7+\sqrt{30}\right)}{2}[/tex]
[tex]=7+\sqrt{30}[/tex]
Similarly, we get,
[tex]x_2=\frac{-\left(-14\right)-2\sqrt{30}}{2\cdot \:1}[/tex]
[tex]x_2=7-\sqrt{30}[/tex]
Therefore the roots of f(x) are 7+root 30 and 7 -root 30.
To learn more about the roots visit:
https://brainly.com/question/2833285
