(1/4)p+(3/4)p=1

Solution for (1/4)p+(3/4)p=1 equation:

(1/4)p+(3/4)p=1

We move all terms to the left:

(1/4)p+(3/4)p-(1)=0


Domain of the equation: 4)p!=0

p!=0/1

p!=0

p∈R


We add all the numbers together, and all the variables

(+1/4)p+(+3/4)p-1=0

We multiply parentheses

p^2+3p^2-1=0

We add all the numbers together, and all the variables

4p^2-1=0

a = 4; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·4·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*4}=\frac{-4}{8}
=-1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*4}=\frac{4}{8}
=1/2
$