+0

# I need help on these...

0
65
4

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
+1968
+1

Hint for #2:

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

Mar 21, 2020
#2
+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

Mar 21, 2020
#3
+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

Mar 21, 2020
#4
0

thanks for all the help!

Mar 22, 2020