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# Help 2!

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1. Find the sum of the cubes of the solutions of x^2-11x+4=0.

2. For integers a, b, c, and d, (x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5. What is the value of a+b+c+d?

Sep 14, 2019

#1
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1.   x^2 - 11x + 4  = 0

Call the solutions a, b

First note, that from Vieta, the sum of the solutions will  =  11/1  = 11

So   a + b  = 11      (1)

Square both sides  of this to give that    a^2 + 2ab + b^2  =  121

Rearrarange this  as   a^2 + b^2  =  121  - 2ab         (2)

And  the product of the roots  =   4/1   = 4

So

ab  =  4     (3)

And the sum of the cubes  =

a^3  + b^3     which we can factor as

(a + b) ( a^2 - ab + b^2)

(a + b) ( [ a^2 + b^2]  - ab )     (4)

Sub (1) , (2) and (3)   into  4  and we have that

a^3 + b^3   =

(11) ( [121 - 2ab] - ab)

(11)  ( 121 - 2(4)  - 4 ]

(11) ( 121 - 12)

(11) ( 109)  =

1199   Sep 14, 2019
#2
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2. For integers a, b, c, and d, (x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5. What is the value of a+b+c+d?

(x^2 + ax + b) ( x^2 + cx +d)   =

x^4 + cx^3  + dx^2

+       ax^3   + (ac)x^2  + (ad)x

+                     bx^2      + (bc)x  + bd

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x^4  + (a + c)x^3  + (ac + d + b)x^2  + ( ad + bc)x  + bd    =  1x^4  + 1x^3 - 2x^2  + 17x -  5

Equating terms   it can be seen that   bd  =  -5

Suppose that   b = -1   and d  = 5

Then

ac + d  + b =   -2      →  ac + 4  =  - 2   →   ac  = -6      (1)

a + c =  1   →  c = 1- a     (2)

Sub (2)  into (1)

a ( 1 - a)  = -6

-a^2 + a  + 6  = 0

a^2 - a - 6  = 0

(a -3) (a + 2)  = 0

So  a  = 3   or a = -2

If we let a  = 3, then c = -2

Then

ad + bc  = 17   ??

3(5) + (-1)(-2)  = 17

15 + 2  = 17    is true

Therfore   a  = 3, b = -1, c = -2  and  d  = 5      and their sum  =  5

BTW.....the same sum is achieved if we let b = 5 and d = -1   Sep 14, 2019
edited by CPhill  Sep 14, 2019