Problem with a quadratic equation and its condition

Question

I've been trying to find a solution to this problem but it seems I cannot find it. Anyway, here is an equation and its condition.

$x^2-2mx+m+1=0$

Condition: $x_1^2 + x_2^2 = 4$

When I get to

$x_1+x_2 = 2m$

$x_1^2 + x_2^2 = 4$

I get stuck.

Answer 1

Since $x_1$ and $x_2$ are solutions to the equation $x^2 -2mx + (m+1) = 0$, we have that

$(x - x_1)(x-x_2) = x^2 - 2mx + (m+1) \\ x^2 - (x_2 + x_1) x + x_1x_2 = x^2 -2mx + (m+1),$

which implies $x_1 + x_2 = 2m$ and $x_1x_2 = m+1$. Therefore,

$x^2_1 + x^2_2 = 4 \\ (x_1 + x_2)^2 - 2x_1x_2 = 4 \\ (2m)^2 - 2(m+1) = 4 \\ 2m^2 - m - 3 = 0 \\ (m + 1)(2m-3) = 0,$

which implies $m_1 = -1$ and $m_2 = 3/2$.