4x-(y-1)^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the linear equation:
4*x-(y-1)^2 = 0
Expand brackets in the left part
4*x-y+1^2 = 0
Looking for similar summands in the left part:
-(-1 + y)^2 + 4*x = 0
Move free summands (without x)
from left part to right part, we given:
4x−(y−1)2+1=1Divide both parts of the equation by (1 - (-1 + y)^2 + 4*x)/x
x = 1 / ((1 - (-1 + y)^2 + 4*x)/x)
We get the answer: x = (-1 + y)^2/4
2 2
im (y) (-1 + re(y)) I*(-1 + re(y))*im(y)
x1 = - ------ + ------------- + --------------------
4 4 2
x1=4(re(y)−1)2+2i(re(y)−1)im(y)−4(im(y))2
x1 = (re(y) - 1)^2/4 + i*(re(y) - 1)*im(y)/2 - im(y)^2/4
Sum and product of roots
[src]
2 2
im (y) (-1 + re(y)) I*(-1 + re(y))*im(y)
- ------ + ------------- + --------------------
4 4 2
4(re(y)−1)2+2i(re(y)−1)im(y)−4(im(y))2
2 2
im (y) (-1 + re(y)) I*(-1 + re(y))*im(y)
- ------ + ------------- + --------------------
4 4 2
4(re(y)−1)2+2i(re(y)−1)im(y)−4(im(y))2
2 2
im (y) (-1 + re(y)) I*(-1 + re(y))*im(y)
- ------ + ------------- + --------------------
4 4 2
4(re(y)−1)2+2i(re(y)−1)im(y)−4(im(y))2
2 2
im (y) (-1 + re(y)) I*(-1 + re(y))*im(y)
- ------ + ------------- + --------------------
4 4 2
4(re(y)−1)2+2i(re(y)−1)im(y)−4(im(y))2
-im(y)^2/4 + (-1 + re(y))^2/4 + i*(-1 + re(y))*im(y)/2