# Which is the sum of the squares of the solutions of the equation 2x^2-mx+1=0?

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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:

2x1^2-m*x1+1=0 (1)

Let's substitute the second root into equation:

2x2^2-m*x2+1=0 (2)

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.

Solution:

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