Question

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

Answer 1

Answer:  $x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}$

Step-by-step 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

$.\ y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}$

Now, let's solve for x:

$xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}$

Lastly, find x² + y² :

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

$x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}$

                                                                                 

$x^2+y^2=\dfrac{21\pm 6\sqrt6}{2}+42\pm 12\sqrt6\\\\\\.\qquad \quad = \dfrac{21\pm 6\sqrt6}{2}+\dfrac{84\pm 24\sqrt6}{2}\\\\\\. \qquad \quad = \dfrac{105\pm 18\sqrt6}{2}$