+2666 Moving everything to the right-hand side gives us: \(x^2 - 3x + 2 = 0 \)
Let a and b be the roots of the quadraitc.
Now, recall the identity: \(a^2 + b^2 = (a+b)^2 - 2ab\)
By Vieta's, we know that \(a + b = -{ b\over a} = 3\) and \(ab = {c \over a} = 2\).
Can you take it from here?
+2666 Moving everything to the right-hand side gives us: \(x^2 - 3x + 2 = 0 \)
Let a and b be the roots of the quadraitc.
Now, recall the identity: \(a^2 + b^2 = (a+b)^2 - 2ab\)
By Vieta's, we know that \(a + b = -{ b\over a} = 3\) and \(ab = {c \over a} = 2\).
Can you take it from here?
