I'm really stuck on this question all comments greatly appreciated ðŸ˜Š
X^2x(2k+2)+5k+1=0

Dazzla16
In order to solve this, we treat it as an algebraic identity to solve for x and k.
x^2  x(2k+2) + 5k + 1 = 0
x^2 2xk 2x + 5k + 1 = 0
x^2 2x = 2xk  5k  1
Since this is an identity, the coefficient of the like terms are equal.
2x = 2xk
2 = 2k
k = 1
Then, substitute k for 1. After that, you can solve for x as normal.
x^2 2x = 2x(1)  5(1) 1
x^2 2x = 2x + 5  1
x^2 = 4
x^2  4 = 0
(x)^2  (2)^2 = 0 (factorising the difference of two squares)
(x + 2)(x  2) = 0
x = 2 or x = 2

Dazzla16
So when k = 1, x = 2 or 2

Dazzla16
In fact, this is not the only solution. If we took a different approach starting from the second section of my solution (from my first comment), we will get another answer.
0k = 5k (there were essentially 0 k's in the other side)
k = 0
Then, substitute k for 0.
x^2  x(2(0)+2) + 5(0) + 1 = 0
x^2 x(2) + 1 = 0
x^2 2x + 1 = 0
(x  1)(x  1) = 0 (factorising the trinomial)
x = 1
Therefore, when k = 0, x = 1
A third solution would be making x^2 = 0x^2, then x = 0
When you work it out, if x = 0, k = 1/5

Me
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