Solve for x
(X-P)^2 =(X- q)^2
To solve the equation (X-P)^2 = (X-q)^2 for X, we can start by expanding both sides of the equation using the concept of perfect square binomials:
(X-P)^2 = (X-q)^2
(X-P)(X-P) = (X-q)(X-q)
Now we can simplify this equation by applying the distributive property:
X^2 - PX - PX + P^2 = X^2 - Xq - Xq + q^2
Combining like terms, we can simplify further:
X^2 - 2PX + P^2 = X^2 - 2Xq + q^2
Next, we can eliminate the X^2 terms from both sides of the equation by subtracting X^2 from both sides:
-2PX + P^2 = -2Xq + q^2
Now, we can bring all the terms involving X to one side of the equation and all the constant terms to the other side:
-2PX + 2Xq = q^2 - P^2
Next, we can factor out X from the left side of the equation:
X(-2P + 2q) = q^2 - P^2
Finally, we can solve for X by dividing both sides of the equation by (-2P + 2q):
X = (q^2 - P^2) / (-2P + 2q)
Therefore, the solution for X is (q^2 - P^2) / (-2P + 2q).
(x-p)^2 = (x-q)^2
just take the root, and you have either
x-p = x-q
so no solution unless p=q, in which case there are infinitely many solutions
OR
x-p = -(x-q)
2x = q+p
x = (q+p)/2
Or, you could have expanded at the start
(x-p)^2 = (x-q)^2
x^2 - 2px + p^2 = x^2 - 2qx + q^2
2(q-p)x = q^2 - p^2 = (q-p)(q+p)
2x = q+p unless q=p
x = (q+p)/2