If x and y are the sum of roots of the quadratic 2x^2 + x -15 = 0
Find
x^2y^3+x^3y^2
i have no idea how to do this. . . .
i need as much help as i can get. . .
thanks =]
2007-08-26 01:19:53 · 6 answers · asked by Hmmmmm 3 in Science & Mathematics ➔ Mathematics
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2007-08-26 01:31:38 · answer #1 · answered by wry humor 5 · 0⤊ 1⤋
The problem is formulated carelessly.
However, if x and y are the roots of above equation, then by the Viete theorem,
xy=-15/2
x+y=-1/2
Therefore,
x^2y^3+x^3y^2
= x^2 y^2 (y+x)
= (xy)^2 (x+y)
= (-15/2)^2 (-1/2)
= -225/8
2007-08-26 08:40:51 · answer #2 · answered by Anonymous · 1⤊ 0⤋
2x^2 + x - 15 = 0
Sum Of Roots (x + y) --> -b/a = -1/2
Product Of Roots (xy) --> c/a = -15/2
x^2y^3 + x^3y^2
= x^2 y^2(y + x)
= (x + y)(xy)^2
= (-1/2)(-15/2)^2
= -28 and 1/8
2007-08-26 08:48:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Factoring it, (2x-5 )(x+ 3)=0.
SO, roots are (5/2) and -3.
So, then x^2y^3+x^3y^2 = x^2y^2(x+y)=
(5/2)^2*(-3)^2*((5/2)-3) = -28.125.
I think thats the way to do it?
2007-08-26 08:39:23 · answer #4 · answered by yljacktt 5 · 1⤊ 0⤋
The roots of the equation you gave:
2x^2 + x -15 = 0
are:
(2x-5)(x+3)=0
x = 5/2 and x = -3
the sum of the roots is 5/2 - 3 = - 0.5
x = y = -.5
(-.5)^2 (-.5)^3 + (-.5)^3 (-.5)^2 = 2(-.5)^5 = -0.0625
2007-08-26 08:38:34 · answer #5 · answered by 037 G 6 · 0⤊ 0⤋
Use the formula x=-b +_ SQRT (b2-4ac) all over 2a : a=2; b=1; c=-15
We get x=2.5; y=3
Substitute in second expression and there we are.
2007-08-26 08:32:16 · answer #6 · answered by galyamike 5 · 0⤊ 1⤋