A root of x2 – 5x – 1 = 0 is
\(x^2-5x-1=0\\\\ a=1,\ b=-5,\ c=-1\\\\ \Delta=b^2-4ac=(-5)^2-4*1*(-1)=25+4=29\\\\ \sqrt{\Delta}=\sqrt{29}\\\\ x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{29}}{2*1}=\frac{5-\sqrt{29}}{2}\\\\ x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{29}}{2*1}=\frac{5+\sqrt{29}}{2}\\\)
Answer:
The roots are \(\frac{5+\sqrt{29}}{2}\) and \(\frac{5-\sqrt{29}}{2}\)
Explanation:
The general form of the quadratic equation is:
ax² + bx + c = 0
The given equation is:
x² - 5x - 1 = 0
By comparison:
a = 1
b = -5
c = -1
To get the roots of the equation, we will use the quadratic formula shown in the attached image.
This means that:
either \(x = \frac{5+\sqrt{(-5)^2-4(1)(-1)}}{2(1)} = \frac{5+\sqrt{29}}{2}\)
or \(x = \frac{5-\sqrt{(-5)^2-4(1)(-1)}}{2(1)} = \frac{5-\sqrt{29}}{2}\)
Hope this helps :)