How to Solve the System of Quadratic Equations
If x² – 5xy + 6y² = 0 and xy – y² = 2, then solve the system of quadratic equations
x^2-5xy+6y^2=0
xy-y^2=2
(x, \ y)=?
Solution to the system of quadratic equations
Let
x² – 5xy + 6y² = 0………………eq(1)
xy – y² = 2………………………….eq(2)
Apply quadratic equation formula in equation 1
x² – 5xy + 6y² = 0
6y² – 5xy + x² = 0
\begin{aligned} y&=\frac {5x \pm \sqrt{(-5x)^2-4 \times 6 \times x^2}}{2 \times 6} \\ \\ &=\frac {5x \pm \sqrt{25x^2-24 x^2}}{12} \\ \\ &=\frac {5x \pm x}{12} \\ \\ \Rightarrow y&=\frac{x}{2} \ \ \& \ \ y=\frac{x}{3}\\ \\ \Rightarrow x&=2y \ \ \& \ \ x=3y\\ \end{aligned}
When x = 2y
From equation 2
xy – y² = 2
⇒ (2y)y – y² = 2
⇒ 2y² – y² = 2
So, y² = 2
⇒ y = ±√2
⇒ x = 2y = ±2√2
When x = 3y
From equation 2
xy – y² = 2
⇒ (3y)y – y² = 2
⇒ 3y² – y² = 2
So, 2y² = 2
y² = 1
⇒ y = ±1
⇒ x = 3y = ±3
So the solution to the system of quadratic equations are (x, y) = (√2, 2√2), (-√2, -2√2), (3, 1) & (-3, -1)