+537 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?
+129852 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
+129852 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
