The solutions to
\(x^2+5x+3=0\)
are p and q and the solutions to
\(x^2+bx+c=0\)
are \(p^2\) and \(q^2\)Find b+c
You can start by finding the solutions to \(x^2+5x+3 \). When you plug that into the quadratic formula, you will end up with \(\frac{-5 \pm \sqrt{13}}{2}\).\(\left(\frac{-5+\sqrt{13}}{2}\right)^2=38-10\sqrt{13}\) and \(\left(\frac{-5-\sqrt{13}}{2}\right)^2=38+10\sqrt{13}\) since those are the solutions to the second quadratic, we know that the second quadratic is the expanded form of \((x-38+10\sqrt{13})(x-38-10\sqrt{13})\). When you expand that, you get \(x^2-76x+144 \). \(b=-76\), and \(c=144 \), so \(b+c=\color{red}{68}\).
x^2 + 5x + 3 = 0
By Vieta
The sum of the roots = -5/1 = -5 = p + q
So
(p + q)^2 = (-5)^2
p^2 + 2pq + q^2 = 25 (1)
And the product of the roots = 3/1 = 3
So
pq = 3
2pq = 6 (2)
Sub (2) into (1) and we get that
p^2 + 6 + q^2 = 25
p^2 + q^2 =19
So
P^2 + q^2 = -b / 1
19 = -b
-19 = b
And, again, by Vieta :
p^2*q^2 = c / 1
(pq)^2 = c
(3)^2 = c
So
b + c =
-19 + 9 =
-10