+0

0
84
3

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

Mar 23, 2020

#1
+8
+1

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 $$b+c=\color{red}{68}$$.

Mar 23, 2020
#2
0

Hmm... I tried that and it's incorrect...

Guest Mar 23, 2020
#3
+111330
+1

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

Mar 23, 2020