Which is the sum of the squares of the solutions of the equation 2x^2-mx+1=0?
Let's recall the fact that a solution of an equation, substituted into equation, verifies it.
Let's substitute the first root into equation and we'll get:
Let's substitute the second root into equation:
If we'll add (1)+(2), we'll get:
2(x1^2 + x2^2) - m*(x1 + x2) + 2 = 0
2(x1^2 + x2^2) = m*(x1 + x2) - 2
x1^2 + x2^2 = [m*(x1 + x2) - 2]/2
We'll find the value of (x1 + x2), using the Viete's relation:
x1+x2 = - (-m)/2
We'll substitute (x1+x2) by m/2
x1^2 + x2^2 = (m*m/2 - 2)/2
x1^2 + x2^2 = (m^2 - 4)/4
To find the sum of the squares of the solutions of the equation 2x^2-mx+1 = 0.
Let 2x^2+mx+1 = (x-x1)(x-x2). Then x1 and x2 are the solutions of the equation.Expanding the right side, we get:
2x^2-mx+1 = x^2-(x1+x2)+x1*x2. If this is an identity, then the coefficients of x^2 , x and constant terms are comparable. Or
2/1 = -m/[-(x1+x2)] = 1/(x1*x2). Or
From the first and 2nd: x1+x2 = m/2 and
from the 1st and 3rd equation, x1*x2 = 1/2
Therefore, x^2+x2^2 = (x1+x2)^2-2x1*x2 = (m/2)^2-2*(1/2) = m^2/4 - 1 = (m^2-4)/4 = (m+2)(m-2)/4