# Quartic Equation with reals

• Sep 28th 2009, 09:29 AM
warriors837
Quartic Equation with reals
If x and y are reals, and x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ?
• Sep 28th 2009, 10:25 AM
Quote:

Originally Posted by warriors837
If x and y are reals, and x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ?

$x^3+y^3=(x+y)(x^2-xy+y^2)=x^2-xy+y^2$

$(x^2+y^2)(x^3+y^3)=(x^2+y^2)(x^2-xy+y^2)=12$

Notice that if we square x+y=1 on both sides we get $x^2+y^2+2xy=1$ Now we have $1+2xy=1$, $2xy=0$. Therefore xy=0. This gives $(x^2+y^2)(x^2-xy+y^2)=(x^2+y^2)^2=12$. So $(x^2+y^2)=\sqrt{12}$
• Sep 28th 2009, 12:19 PM
warriors837
No, I got 3.
• Sep 28th 2009, 12:33 PM
warriors837
Since, x+y=1, then we can replaced y with 1-x. Thus, the result or x^2 + y^2 will be 3.

See attachment....

Do you agree?
• Sep 28th 2009, 04:54 PM
pacman
Given: "x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ? "

Solution: since (x + y) = 1, we can generate identities from it;

(x + y)^2 = x^2 + 2xy + y^2 = 1

x^2 + y^2 = 1 - 2xy - - - (1)

(x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 = 1

x^3 + y^3 +3xy(x + y) = 1

x^3 + y^3 + 3xy(1) = 1

x^3 + y^3 = 1 - 3xy - - - (2)

From the given expression: (x^2 + y^2)(x^3 + y^3) = 12, we equate (1) and (2)

we have (1 - 2xy)(1 - 3xy) = 12, expanding by FOIL Method

1 - 5xy + 6(xy)^2 = 12, rearranging

6(xy)^2 - 5xy - 11 = 0, by quadratic formula,

xy = [5 +/- (25 - 4(6)9(-11))^0.5]/2(6)

roots for xy: xy = (5 +/- 17)/12

(+) xy = 22/12 or

(-) xy = -12/12

for convenience SAY, xy = -1

then from (1)

x^2 + y^2 = 1 - 2xy = 1 - 2(-1) = 3.

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• Sep 28th 2009, 05:02 PM
pacman