All Leaving Cert Mathematics posts
• #### Maths help! Whitefox

I'm really stuck on this question all comments greatly appreciated ðŸ˜Š

X^2-x(2k+2)+5k+1=0

1. #### 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

2. #### Dazzla16

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

3. #### 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