Solve quadratic equations using the quadratic formula from the reference sheet, and use the discriminant to determine the number of real solutions (LA A1: A-REI.B.4).

What this topic is asking

Standard A1: A-REI.B.4 includes the quadratic formula, the universal solver that is printed on the LEAP mathematics reference sheet, so the credit is for substituting correctly and simplifying. The discriminant (the part under the radical) gives the number of real solutions without fully solving, a frequent quick item. These are Type I Major Content items.

Using the quadratic formula

Set the equation to $ax^2 + bx + c = 0$, identify $a$, $b$, $c$, then substitute.

The two error-prone spots are the sign of $-b$ and the sign of $-4ac$. When $c$ is negative, $-4ac$ becomes positive, increasing the discriminant.

The discriminant: counting real solutions

The discriminant is $b^2 - 4ac$, the expression under the radical. Its sign tells you how many real solutions exist, which matches how many times the parabola meets the $x$-axis.

• $b^2 - 4ac > 0$: two real solutions (the parabola crosses the axis twice).
• $b^2 - 4ac = 0$: one real solution, a double root (the vertex sits on the axis).
• $b^2 - 4ac < 0$: no real solutions (the parabola misses the axis).

For $x^2 + 2x + 5 = 0$: $b^2 - 4ac = 4 - 20 = -16 < 0$, so no real solutions. Computing the discriminant is much faster than solving when only the count is asked.

How LEAP examines this topic

• Equation response. Solve with the formula and enter the solutions in simplest radical form.
• Multiple choice. Count real solutions from the discriminant, with sign-error distractors.
• Inline choice. State the number of solutions and whether the parabola crosses the axis.

A clarifying idea: the discriminant lives inside the formula, so the formula and the count are the same computation: if the discriminant is negative, the square root of a negative has no real value, which is exactly why there are no real solutions.

Why the quadratic formula always works

The quadratic formula solves every quadratic because it is derived by completing the square on the general equation $ax^2 + bx + c = 0$, with no assumptions about whether the numbers are nice. That derivation is why the formula is a safe fallback: if a quick check shows no integer factor pair multiplies to $ac$ and adds to $b$, the quadratic does not factor over the integers, and hunting for a factorization that does not exist wastes time, go straight to the formula. The trade-off is more arithmetic, so reserve it for quadratics that resist factoring and keep factoring or square roots for the cases where they are quicker. The discriminant's role falls straight out of the derivation: it is whatever ends up under the square root, and a square root behaves differently on positives, zero, and negatives, two real values from a positive (the $\pm$ splits them), one from zero (the $\pm$ collapses), none from a negative (no real root). On the calculator sessions you may evaluate the formula numerically, but exact-match items still expect simplest radical form, so reduce $\sqrt{40}$ to $2\sqrt{10}$ rather than entering a decimal unless a rounded value is requested.

Connecting the discriminant to the graph

The discriminant has a clean graphical meaning the test likes to probe. Because the solutions are the $x$-intercepts of $y = ax^2 + bx + c$, a positive discriminant means the parabola crosses the $x$-axis at two distinct points, a zero discriminant means it touches the axis at exactly one point (the vertex is on the axis), and a negative discriminant means the parabola never reaches the axis. So reading the discriminant tells you the shape of the picture, not just a count, which is why an item may ask both "how many real solutions?" and "does the graph cross the $x$-axis?" in one breath.

Try this

Q1. Solve $x^2 + 4x + 1 = 0$ in simplest radical form. [2 points]

• Cue. $x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}$.

Q2. How many real solutions does $x^2 - 6x + 9 = 0$ have? [1 point]

• Cue. $b^2 - 4ac = 36 - 36 = 0$, so one (double) solution.