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1. 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.

 

2. Let a and b be the solutions to 7x^2 + x - 5 = 0. Compute (a - 4)(b - 4).

 Mar 21, 2020
 #1
avatar+1968 
+1

Hint for #2: 

 

Factor the equation and let a and b be the roots of it. Then, compute!

 Mar 21, 2020
 #2
avatar+111329 
+1

1. 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.

 

By Vieta

 

p + q  =   -5/1   =  -  5

 

Squaring  we  have that

 

p^2 + 2pq + q^2  =  25

 

And

 

pq  = 3/1   =  3      ....so....

2pq  = 6

 

Which implies  that

 

p^2  + 6  + q^2  = 25

 

p^2  + q^2  =  19

 

Which  means that in x^2 + bx +c 

 

-b/1   = p^2 + q^2

 

-b  =  19

 

b =   -19

 

And

 

p^2q^2  =  c /1

 

(pq)^2  =  c

 

(3)^2  = c

 

9 =  c

 

So

 

b + c  =   -19  + 9    =    -10

 

 

 

cool cool cool

 Mar 21, 2020
 #3
avatar+111329 
+1

2. Let a and b be the solutions to 7x^2 + x - 5 = 0. Compute (a - 4)(b - 4).

 

Again, by Vieta

 

a + b  =  -1/7      

 

ab   =  -5/7

 

So

 

(a - 4) ( b - 4)   =

 

ab  - 4a - 4b  + 16   = 

 

ab  -4 (a + b)  + 16    =

 

-5/7  - 4(-1/7)  + 16   =

 

-1/7  +  16  =

 

-1/7  +  112/7  =

 

111 / 7

 

 

 

cool cool cool

 Mar 21, 2020
 #4
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0

thanks for all the help!

 Mar 22, 2020

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