Question

P^2-4p=1
step by step explanation pleaseee

Answer 1

`p_1=2+ √(5)\\p_2=2- √(5)`

Explanation:

A second-degree equation is often expressed as:

`ax^2+bx+c=0`

where a,b, and c are constants.

Solving with the quadratic formula:

`\displaystyle x=(-b\pm √(b^2-4ac))/(2a)`

The equation to solve is:

`p^2-4p=1`

We need to express the equation as mentioned above. The right side of the equation must be zero, so we subtract 1:

`p^2-4p-1=0`

Now we have the values to solve the equation:

a=1, b=-4, c=-1

Applying the formula:

`\displaystyle p=(-(-4)\pm √((-4)^2-4(1)(-1)))/(2(1))`

`\displaystyle p=(4\pm √(16+4))/(2)`

`\displaystyle p=(4\pm √(20))/(2)`

Since 20=4*5:

`\displaystyle p=(4\pm √(4*5))/(2)`

Taking the square root of 4:

`\displaystyle p=(4\pm 2√(5))/(2)`

Simplifying by 2:

`p=2\pm √(5)`

Two solutions are found:

`\boxed{p_1=2+ √(5)}\\\boxed{p_2=2- √(5)}`