Question

If x+y=12 and xy=15,find the value of (x^2+y^2)

Answer 1

Answer:
`x^2+y^2=(105\pm 18\sqrt6)/(2)`

Explanation:

EQ1: x + y = 12 --> x = 12 - y

EQ2: xy = 15

Substitute x = 12-y into EQ2 to solve for y:

(12 - y)y = 15

12y - y² = 15

0 = y² - 12y + 15

↓ ↓ ↓

a=1 b= -12 c=15

`.\ y=(-b\pm √(b^2-4ac))/(2a)\\\\\\.\quad =(-(-12)\pm √((-12)^2-4(1)(15)))/(2(1))\\\\\\.\quad =(12\pm √(144-120))/(2)\\\\\\.\quad =(12\pm √(24))/(2)\\\\\\.\quad =(12\pm 2√(6))/(2)\\\\\\.\quad =6\pm √(6)`

Now, let's solve for x:

`xy=15\\\\x(6\pm\sqrt6)=15\\\\x=(15)/(6\pm\sqrt6)\\\\\\x=(15)/(6\pm\sqrt6)\bigg((6\pm\sqrt6)/(6\pm\sqrt6)\bigg)=(6\pm \sqrt6)/(2)`

Lastly, find x² + y² :

`y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6`

`x^2=\bigg((6\pm \sqrt6)/(2)\bigg)^2\quad \rightarrow \quad x^2=(42\pm 12\sqrt6)/(4)\quad \rightarrow \quad x^2=(21\pm 6\sqrt6)/(2)`

`x^2+y^2=(21\pm 6\sqrt6)/(2)+42\pm 12\sqrt6\\\\\\.\qquad \quad = (21\pm 6\sqrt6)/(2)+(84\pm 24\sqrt6)/(2)\\\\\\. \qquad \quad = (105\pm 18\sqrt6)/(2)`