Non Linear Equations in One Variable and System of Equations in Two Variables Pattern - Number of Solutions

Digital SAT® Math — Non Linear Equations in One Variable and System of Equations in Two Variables

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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.

Learn the pattern. Then lock it in.

The SAT repeats question patterns. Miss them, and you lose points. Recognize them fast, and you gain points. JustLockedIn shows you which patterns are hurting your score and gives you focused practice to fix them.

Practice this pattern → 110 practice questions available

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