Question

Solve the system of two quadratic equations with two variables

\begin{cases} xy+x+y=-13 \qquad\qquad (1)\\x^2+y^2 = 29 \qquad\qquad\qquad\: (2)\end{cases}

Collected in the board: System of quadratic equations

Steven Zheng posted 1 week ago

Answer

Observe the equations and add 2\times ( 1)+(2)

(x+y)^2+2(x+y)-3=0

Factoring the LHS

(x+y-3)(x+y+1) = 0

Then, we get

x+y-3=0 \text{ or } x+y+1=0

Substituting them to (1) gives two system of equations

\begin{cases} x+y-3=0 \\xy = -10 \end{cases}
(3)

\qquad \text{or}\qquad

\begin{cases} x+y+1=0 \\xy=14 \end{cases}
(4)

Solving (3) obtains

\begin{cases} x_1=-5 \\y_1=2 \end{cases} \qquad \text{or}\qquad\begin{cases} x_2=-2 \\y_2=-5 \end{cases}


Solving (4) obtains

\begin{cases} x_3=\dfrac{1+\sqrt{57} }{2} \\y_3=\dfrac{1-\sqrt{57} }{2} \end{cases} \qquad \text{or}\qquad \begin{cases} x_4=-\dfrac{1-\sqrt{57} }{2} \\y_4=\dfrac{1+\sqrt{57} }{2} \end{cases}

Steven Zheng posted 1 week ago

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