Answer :
[tex]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}\\
[/tex]
Answer:
The roots are [tex] \frac{5+\sqrt{29}}{2} [/tex] and [tex] \frac{5-\sqrt{29}}{2} [/tex]
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 [tex] x = \frac{5+\sqrt{(-5)^2-4(1)(-1)}}{2(1)} = \frac{5+\sqrt{29}}{2} [/tex]
or [tex] x = \frac{5-\sqrt{(-5)^2-4(1)(-1)}}{2(1)} = \frac{5-\sqrt{29}}{2} [/tex]
Hope this helps :)