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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
avatar+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
avatar
0

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

Guest Mar 23, 2020
 #3
avatar+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

 

 

 

cool cool cool

 Mar 23, 2020

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