Problem 9

Find all real solutions \((x,y)\) to \(5x^2 - 2xy + 2y^2 - 2x - 2y + 1 = 0\).

Slide Notes

If \(x=0\), the equation becomes

\[2y^2 - 2y + 1 = 0\]

The discriminant is

\[\begin{align*} \Delta & = (-2)^2 - 4(2)(1) \\ & = -4 \\ & \lt 0 \end{align*}\]

and so this quadratic equation has no real roots.

If \(x=1\), the equation becomes

\[2y^2 - 4y + 4 = 0\]

The discriminant is

\[\begin{align*} \Delta & = (-4)^2 - 4(2)(4) \\ & = -16 \\ & \lt 0 \end{align*}\]

and so this quadratic equation has no real roots.

If \(x=-1\), the equation becomes

\[2y^2 + 8 = 0\]

The equation \(2y^2=-8\) has no real solutions.

If \(x=2\), the equation becomes

\[2y^2 -6y + 17 = 0\]

The discriminant is again negative.

If \(x=-2\), the equation becomes

\[2y^2 +2y + 25 = 0\]

The discriminant is again negative.

This isn't leading anywhere.

Does this mean that there are no solutions?

Hint 1: Try writing as a quadratic equation in \(x\). Some or all of the coefficients will be in terms of \(y\). Does this help?

Hint 2: Try solving for \(x\) in terms of \(y\) using the quadratic formula.

We write the equation as \(5x^2 - x(2y+2) + (2y^2-2y+1) = 0\).

\(x\)

\(= \dfrac{(2y+2) \pm \sqrt{(2y+2)^2 - 4(5)(2y^2-2y+1)}}{2(5)}\)

\(= \dfrac{2y+2 \pm \sqrt{4y^2+8y+4-40y^2+40y-20}}{10}\)

\(= \dfrac{2y+2 \pm \sqrt{-36y^2 + 48y - 16}}{10}\)

\(= \dfrac{y+1 \pm \sqrt{-(9y^2 -12y +4)}}{5}\)

\(= \dfrac{y+1 \pm \sqrt{-(3y-2)^2}}{5}\)

For \(x\) to be real, we need \(-(3y-2)^2 \geq 0\). But \(-(3y-2)^2 \leq 0\).

This means that \(3y-2=0\), or \(y = \dfrac{2}{3}\), which gives \(x = \dfrac{\frac{2}{3}+1\pm 0}{5}\)\(= \dfrac{1}{3}\).

Thus, \((x,y) = \left(\frac{1}{3},\frac{2}{3}\right)\) is the only solution.

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