Thread: 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 ?

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 ?

Notice that if we square x+y=1 on both sides we get Now we have , . Therefore xy=0. This gives . So

No, I got 3.

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?

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.

----------------------------------------------

-

warriors837, my solution confirms your answer,

your quartic equation 6x^4 - 12x^3 + 11x^2 - 5x + 1 = 12 has indeed two REAL roots, x1 = 1/2 (1 - sqrt(5)) and x2 = 1/2 (1 + sqrt(5))

graph below was generated, see

Could you explain your approach more in depth?

warriors837, that's the most elaborate solution i made in this forum. I can't make it simplier . . .. after all we have the same answer.

my answer hinges on the the value of xy, please reread my solution, it utilizes a binomial square and cube.