Answer:
[tex]p_1=2+ \sqrt{5}\\p_2=2- \sqrt{5}[/tex]
Step-by-step explanation:
A second-degree equation is often expressed as:
[tex]ax^2+bx+c=0[/tex]
where a,b, and c are constants.
Solving with the quadratic formula:
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The equation to solve is:
[tex]p^2-4p=1[/tex]
We need to express the equation as mentioned above. The right side of the equation must be zero, so we subtract 1:
[tex]p^2-4p-1=0[/tex]
Now we have the values to solve the equation:
a=1, b=-4, c=-1
Applying the formula:
[tex]\displaystyle p=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(-1)}}{2(1)}[/tex]
[tex]\displaystyle p=\frac{4\pm \sqrt{16+4}}{2}[/tex]
[tex]\displaystyle p=\frac{4\pm \sqrt{20}}{2}[/tex]
Since 20=4*5:
[tex]\displaystyle p=\frac{4\pm \sqrt{4*5}}{2}[/tex]
Taking the square root of 4:
[tex]\displaystyle p=\frac{4\pm 2\sqrt{5}}{2}[/tex]
Simplifying by 2:
[tex]p=2\pm \sqrt{5}[/tex]
Two solutions are found:
[tex]\boxed{p_1=2+ \sqrt{5}}\\\boxed{p_2=2- \sqrt{5}}[/tex]
