Quadratics with no real solutions

The discriminant $\Delta = b^2 - 4ac$ tells us how many real solutions a quadratic equation $ax^2 + bx + c = 0$ has. We already know the three cases: $\Delta > 0$ gives two real solutions, $\Delta = 0$ gives one repeated solution, and $\Delta < 0$ gives none. But why does a negative discriminant rule out real solutions entirely?

Look at the quadratic formula:

When $\Delta < 0$, the formula asks us to take the square root of a negative number. No real number squared gives a negative result: positive numbers square to positives, negative numbers square to positives, and $0^2 = 0$. So $\sqrt{\Delta}$ does not exist in the real numbers, and neither does $x$.

Sometimes we can spot this without computing $\Delta$ at all. For instance, $x^2 + 9 = 0$ rearranges to $x^2 = -9$. Since no real number squares to $-9$, there are no real solutions.

Determine whether $2x^2 - 3x + 5 = 0$ has real solutions.

Solution

We identify $a = 2$, $b = -3$, and $c = 5$, then compute the discriminant:

Since $\Delta = -31 < 0$, the quadratic formula would require $\sqrt{-31}$, which has no real value. The equation has no real solutions.

Every quadratic $y = ax^2 + bx + c$ graphs as a U-shaped curve called a parabola. The real solutions of $ax^2 + bx + c = 0$ are the $x$-coordinates where the parabola crosses the $x$-axis (where $y = 0$). The discriminant tells us how many crossings there are:

• $\Delta > 0$: the parabola crosses the $x$-axis twice (two real solutions).
• $\Delta = 0$: the parabola just touches the $x$-axis at its turning point (one repeated solution).
• $\Delta < 0$: the parabola misses the $x$-axis entirely (no real solutions).

The turning point of a parabola is its lowest or highest point. The sign of $a$ determines the direction:

• $a > 0$: the parabola opens upward (U-shape), so the turning point is the lowest point. If this lowest point sits above the $x$-axis, the parabola never reaches $y = 0$ - no real solutions.
• $a < 0$: the parabola opens downward (inverted U), so the turning point is the highest point. If this highest point sits below the $x$-axis, the parabola again never reaches $y = 0$.

Use the sliders to adjust $a$, $b$, and $c$ and watch how the discriminant and the number of $x$-axis crossings change together.

Does $x^2 - 2x + 5 = 0$ have real solutions? What does the parabola $y = x^2 - 2x + 5$ look like?

Solution

First, the discriminant:

So the equation has no real solutions. Geometrically, this means the parabola does not cross the $x$-axis. We can see why in the graph below:

The parabola opens upward ($a = 1 > 0$), so its turning point is the lowest point on the curve. That lowest point is at $y = 4$, which is above the $x$-axis - the dashed line shows the gap. The parabola never comes down far enough to reach $y = 0$, so there are no $x$-axis crossings and no real solutions.