Question

Найдите сумму квадратов корней x^2-7x+5=0

Answer 1

$x^2-7x+5=0 \\ D = (-7)^2 - 4*5 = 49 - 20 = 29 \\ x_{1} = \frac{7+ \sqrt{29}}{2}$
$x_{2} = \frac{7- \sqrt{29}}{2}$ 
$x^2_{1} + x^2_{2} = (\frac{7+ \sqrt{29}}{2})^2 +(\frac{7- \sqrt{29}}{2})^2 = \frac{78+14 \sqrt{29} }{4}+\frac{78-14 \sqrt{29} }{4} = \frac{78+78}{4} = 39$