Review question

# When do these simultaneous equations have no real solutions? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9658

## Solution

From the equations \begin{align*} 3x + 2y &= 5, \\ 3x^2 + 2y^2 &= k \end{align*}

derive an equation relating $x$ and $k$.

By rearranging the first equation, we have that $y = \frac{5 - 3x}{2}.$ By substituting this into the second equation, we see that \begin{align*} 3x^2 + 2 \left( \frac{5 - 3x}{2} \right)^2 = k &\iff 3x^2 + \frac{(5 - 3x)^2}{2} = k \\ &\iff 3x^2 + \frac{25}{2} - 15x + \frac{9}{2} x^2 = k \\ &\iff \frac{15}{2} x^2 - 15x + \frac{25}{2} = k\\ &\iff 15x^2 - 30x + 25 = 2k. \end{align*}

Show that there are no real values of $x$ if $k < 5$.

Recall that for a quadratic equation $ax^2 + bx + c = 0$, the discriminant is the expression $b^2 - 4ac$, and that the quadratic has no real solutions if and only if $\Delta < 0$.

The previous part of the question led to the quadratic equation $15x^2-30x+25=2k$, which we can rewrite as $15x^2-30x+(25-2k)=0$.

The associated discriminant is then \begin{align*} (-30)^2 - 4 \times 15(25-2k) &= 900-60(25-2k)\\ &= -600+120k. \end{align*}

Therefore there are no real values of $x$ if and only if $-600+120k<0$, or $120k<600$. Dividing by $120$ gives us $k<5$ as required.

Geometrically, the two equations represent a fixed straight line and an ellipse of size determined by $k$. When the ellipse is small enough, the line will not intersect it, as can be seen in this interactivity.