Number of Solutions

Number of Solutions (Discriminant)

This pattern asks how many real solutions a quadratic equation has — or what value of a constant produces a specific number of solutions. The tool is the discriminant: for $ax^2 + bx + c = 0$, the discriminant is $\Delta = b^2 - 4ac$.

$\Delta > 0$: two distinct real solutions
$\Delta = 0$: exactly one real solution (a repeated root)
$\Delta < 0$: no real solutions

Worked Examples

Example 1. $2x^2 + 5x - 3 = 0$. How many distinct real solutions does this equation have?

A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many

Identify $a = 2$, $b = 5$, $c = -3$.
$\Delta = 5^2 - 4(2)(-3) = 25 + 24 = 49$
Since $49 > 0$, there are two distinct real solutions.
Gotcha: Be careful with the sign of $c$. Here $c = -3$, so $-4ac = -4(2)(-3) = +24$. The double negative makes it positive.
The answer is C.

Example 2. $4x^2 + kx + 49 = 0$. For what value of $k$ does this equation have two distinct real solutions?

A) $28$
B) $15$
C) $-20$
D) $30$

For two distinct solutions: $\Delta > 0$
$k^2 - 4(4)(49) > 0$
$k^2 - 784 > 0$
$k^2 > 784$
$|k| > 28$
Check the options: $k = 28$ gives $\Delta = 0$ (exactly one solution), $k = 15$ and $k = -20$ give $\Delta < 0$ (no solutions). Only $k = 30$ gives $|30| = 30 > 28$, so $\Delta > 0$.
Gotcha: $k = 28$ gives $\Delta = 0$, which means exactly one solution, not two. The boundary value doesn't count for "two distinct solutions" — you need $\Delta$ strictly greater than zero.
The answer is D.

Example 3. $25x^2 - 30x + k = 0$. The equation has exactly one distinct real solution. What is $k$?

For exactly one solution: $\Delta = 0$
$(-30)^2 - 4(25)(k) = 0$
$900 - 100k = 0$
$k = 9$
Gotcha: Make sure you square the entire coefficient $b = -30$: $(-30)^2 = 900$, not $-900$. The discriminant formula uses $b^2$, which is always positive.
The answer is $k = 9$.

Example 4. A horizontal line $4y = k$ is tangent to the parabola $y = -x^2 + 6x$. What is $k$?

"Tangent" means the line touches the parabola at exactly one point → exactly one solution → $\Delta = 0$.
The line is $y = \dfrac{k}{4}$. Set equal to the parabola:
$\dfrac{k}{4} = -x^2 + 6x$
$x^2 - 6x + \dfrac{k}{4} = 0$
$\Delta = (-6)^2 - 4(1)\left(\dfrac{k}{4}\right) = 36 - k = 0$
$k = 36$
Gotcha: The vertex of $y = -x^2 + 6x$ is at $x = 3$, giving $y = 9$. A horizontal line tangent to a downward parabola touches it at the vertex. Check: $4(9) = 36$ ✓.
The answer is $k = 36$.

Example 5. Which of the following equations has no real solutions?

When you see this question type, compute the discriminant for each option. The one with $\Delta < 0$ has no real solutions.
Example: $3x^2 + 2x + 5 = 0$ → $\Delta = 4 - 60 = -56 < 0$ → no real solutions ✓
vs. $3x^2 + 2x - 5 = 0$ → $\Delta = 4 + 60 = 64 > 0$ → two solutions ✗

Example 6. $x^2 + 6x + c = 0$ has exactly one solution. What is $c$?

$\Delta = 36 - 4c = 0$, so $c = 9$.
Notice: $x^2 + 6x + 9 = (x + 3)^2$ — a perfect square. When the discriminant is zero, the quadratic is always a perfect square.

What to Do on Test Day

• Discriminant formula: $\Delta = b^2 - 4ac$. Memorize it. The SAT tests it heavily.
• Interpretation: $\Delta > 0$ → two solutions. $\Delta = 0$ → one solution. $\Delta < 0$ → no solutions.
• "Tangent to" = one intersection → set $\Delta = 0$.
• Sign traps in $b^2$: $b^2$ is always positive, even when $b$ is negative. $(-30)^2 = 900$.
• Sign traps in $-4ac$: When $a$ and $c$ have the same sign, $-4ac$ is negative (reduces the discriminant). When they have opposite signs, $-4ac$ is positive (increases it).
• Boundary check: If the question says "two distinct solutions," $\Delta$ must be strictly $> 0$, not $= 0$. The value that makes $\Delta = 0$ gives exactly one, not two.
• Perfect squares: When $\Delta = 0$, the quadratic factors as $(px + q)^2$. This can be a useful shortcut — if the quadratic looks like a perfect square, the discriminant is zero.