Hi all,
I am just new to this site and it looks like such an awesome resource.
Can anyone help me with this???
If x + y = 1 and x³ + y³ = 19, find the value of x² + y²
Hello, Joel!
Welcome aboard!
I have a back-door approach . . .
If .$\displaystyle \begin{array}{cccc}x + y &=& 1 & {\color{blue}[1]} \\ x^3 + y^3 &=& 19 & {\color{blue}[2]} \end{array}$ . find the value of $\displaystyle x^2+y^2.$
Cube [1]: .$\displaystyle (x+y)^3 \:=\:1^3\quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1$
$\displaystyle \text{We have: }\;\underbrace{(x^3 + y^3)}_{\text{This is 19}} + \;3xy\underbrace{(x+y)}_{\text{This is 1}} \:=\:1 \quad\Rightarrow\quad 19 \;+ \;3xy \:=\:1 \quad\Rightarrow\quad xy \:=\:-6$
Square [1]: .$\displaystyle (x+y)^2 \:=\:1^2 \quad\Rightarrow\quad x^2+2xy + y^2 \:=\:1$
$\displaystyle \text{We have: }\;x^2 + y^2 + 2(xy) \:=\:1$
. . . . . . . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . . . $\displaystyle ^{\text{This is }\text{-}6}$
Therefore: .$\displaystyle x^2+y^2 - 12 \:=\:1 \quad\Rightarrow\quad \boxed{x^2+y^2 \:=\:13}$
This method is guarenteed to impress/surprise/terrify your teacher.
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