Question

se xy=27 e (x-y)=10, qual é o valor da expressão (x²-y²)? cálculo coerente.

Answer 1

$\begin{cases}xy=27\\x-y=10\longrightarrow{x}=10+y\end{cases}\\\\\\xy=27\longrightarrow(10+y)y=27\longrightarrow10y+y^2=27\longrightarrow{y}^2+10y-27=0\longrightarrow\\\\y=\dfrac{-(10)\pm\sqrt{(10)^2-4\cdot{(1)}\cdot{(-27)}}}{2\cdot(1)}\longrightarrow{y}=\dfrac{-10\pm\sqrt{208}}{2}\longrightarrow\\\\\\y=\dfrac{-10\pm4\sqrt{13}}{2}\left\langle\begin{matrix}y'=-5+2\sqrt{13}\\\\y''=-5-2\sqrt{13}\end{matrix}\right\\\\\\\\\begin{matrix}\text{Para}\,\,y':\\\\x=10+y\\\\x=10+(-5+2\sqrt{13})\\\\x'=5+2\sqrt{13}\end{matrix}\qquad\qquad\qquad\begin{matrix}\text{Para}\,\,y'':\\\\x=10+y\\\\x=10+(-5-2\sqrt{13})\\\\x''=5-2\sqrt{13}\end{matrix}\\\\\\\\\begin{matrix}\text{Para}\ (x',\,y'):\\\\x^2-y^2=(x+y)\cdot(x-y)\\\\x^2-y^2=[(5+2\sqrt{13})+(-5+2\sqrt{13})]\cdot[(5+2\sqrt{13})-(-5+2\sqrt{13})]\\\\x^2-y^2=[4\sqrt{13}]\cdot[10]\\\\\boxed{x^2-y^2=40\sqrt{13}}\end{matrix}\\\\\\\\\begin{matrix}\text{Para}\,\,(x'',\,y''):\\\\x^2-y^2=(x+y)\cdot(x-y)\\\\x^2-y^2=[(5-2\sqrt{13}+-5-2\sqrt{13})]\cdot[(5-2\sqrt{13})-(-5-2\sqrt{13})]\\\\x^2-y^2=[-4\sqrt{13}]\cdot[10]\\\\\boxed{x^2-y^2=-40\sqrt{13}}\end{matrix}$