AlgebraDifficulty 15-6533 tagged problems

Quadratic Discriminant

Quadratic Discriminant is the habit of using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\Delta = b^2 - 4ac \quad\text{for}\quad ax^2 + bx + c = 0,\ a \ne 0.$$

The sign of $b^2 - 4ac$ tells you whether a quadratic has two real roots, one repeated root, or no real roots — and when it is a perfect square the roots are rational.

Why It Works

1
The quadratic formula gives $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, so the term under the radical is $\Delta = b^2 - 4ac$.
2
If $\Delta > 0$ the square root is a positive real and there are two distinct real roots; if $\Delta = 0$ both roots coincide at $x = -\frac{b}{2a}$.
3
If $\Delta < 0$ the square root is imaginary, so there are no real roots (two complex conjugates instead).
4
When $a, b, c$ are integers and $\Delta$ is a perfect square, $\sqrt{\Delta}$ is an integer, so the roots are rational.

Worked Examples

Example 1

For which value of $k$ does $x^2 + kx + 9 = 0$ have exactly one (repeated) real root?

1
A repeated root occurs exactly when the discriminant is zero.
2
Here $a = 1$, $b = k$, $c = 9$, so $\Delta = k^2 - 4(1)(9) = k^2 - 36$.
3
Set $\Delta = 0$: $k^2 = 36$, so $k = 6$ or $k = -6$.

Answer:

$k = \pm 6$.

Warm-up

How many real roots does $2x^2 + 3x + 5 = 0$ have?

1
Identify $a = 2$, $b = 3$, $c = 5$.
2
Compute $\Delta = b^2 - 4ac = 3^2 - 4(2)(5) = 9 - 40 = -31$.
3
Since $\Delta < 0$, there are no real roots (the two roots are complex conjugates).

Answer:

No real roots ($\Delta = -31 < 0$).

Contest level

For what value of $c$ is the line $y = 2x + c$ tangent to the parabola $y = x^2$?

1
Tangency means the line meets the parabola in exactly one point, so $x^2 = 2x + c$ has a repeated root.
2
Rearrange: $x^2 - 2x - c = 0$, with $a = 1$, $b = -2$, constant $-c$.
3
Set $\Delta = 0$: $(-2)^2 - 4(1)(-c) = 4 + 4c = 0$, so $c = -1$.

Answer:

$c = -1$ (tangent at $x = 1$).

Olympiad / Challenge

Find all integers $n$ for which $x^2 + nx + (n + 3) = 0$ has integer roots.

1
Integer roots require $\Delta = n^2 - 4(n + 3) = n^2 - 4n - 12$ to be a perfect square, say $k^2$ with $k \ge 0$. Complete the square: $(n - 2)^2 - 16 = k^2$.
2
Rearrange to a difference of squares: $(n - 2)^2 - k^2 = 16$, i.e. $(n - 2 - k)(n - 2 + k) = 16$. The two factors have equal sums to $2(n-2)$, so they share parity; both must be even.
3
Same-parity factor pairs of $16$ (with $n - 2 - k \le n - 2 + k$) are $(2, 8), (4, 4), (-8, -2), (-4, -4)$, giving $n - 2 \in \{5, 4, -5, -4\}$, so $n \in \{7, 6, -3, -2\}$.
4
Each makes $\Delta$ a perfect square AND the roots $\frac{-n \pm k}{2}$ integers: $n=7 \Rightarrow (x+2)(x+5)$; $n=6 \Rightarrow (x+3)^2$; $n=-3 \Rightarrow x(x-3)$; $n=-2 \Rightarrow (x-1)^2$.

Answer:

$n \in \{-3, -2, 6, 7\}$.

Going Deeper

Generalization: for any polynomial the discriminant (a symmetric function of the coefficients) vanishes exactly when two roots coincide; for the cubic $x^3 + px + q$ it is $-4p^3 - 27q^2$.
Where it appears: AMC 'for how many $k$ are the roots real/rational' counting, AIME problems forcing $\Delta$ to be a perfect square for integer roots, and tangency conditions (line meets conic once $\iff \Delta = 0$).
Pitfall: $\Delta$ being a perfect square only guarantees RATIONAL roots when $a, b, c$ are integers; with non-integer coefficients that fails. Also $\Delta \ge 0$ (not $> 0$) is the condition for 'at least one real root' — do not exclude the repeated-root boundary $\Delta = 0$.

Spot the Signal

Look for problems where the key step is using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges.
You can describe the hard part as using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for quadratic discriminant, then rewrite the givens around it.
Name the relation that makes Quadratic Discriminant legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Quadratic Discriminant just because the surface looks familiar; verify the required condition first.
Applying Quadratic Discriminant because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let quadratic discriminant reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges.

Back to techniques

AlgebraDifficulty 15-6533 tagged problems

Quadratic Discriminant

Practice this technique Match my level

The Key Result

$$\Delta = b^2 - 4ac \quad\text{for}\quad ax^2 + bx + c = 0,\ a \ne 0.$$

The sign of $b^2 - 4ac$ tells you whether a quadratic has two real roots, one repeated root, or no real roots — and when it is a perfect square the roots are rational.

Why It Works

1
The quadratic formula gives $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, so the term under the radical is $\Delta = b^2 - 4ac$.
2
If $\Delta > 0$ the square root is a positive real and there are two distinct real roots; if $\Delta = 0$ both roots coincide at $x = -\frac{b}{2a}$.
3
If $\Delta < 0$ the square root is imaginary, so there are no real roots (two complex conjugates instead).
4
When $a, b, c$ are integers and $\Delta$ is a perfect square, $\sqrt{\Delta}$ is an integer, so the roots are rational.

Worked Examples

Example 1

For which value of $k$ does $x^2 + kx + 9 = 0$ have exactly one (repeated) real root?

1
A repeated root occurs exactly when the discriminant is zero.
2
Here $a = 1$, $b = k$, $c = 9$, so $\Delta = k^2 - 4(1)(9) = k^2 - 36$.
3
Set $\Delta = 0$: $k^2 = 36$, so $k = 6$ or $k = -6$.

Answer:

$k = \pm 6$.

Warm-up

How many real roots does $2x^2 + 3x + 5 = 0$ have?

1
Identify $a = 2$, $b = 3$, $c = 5$.
2
Compute $\Delta = b^2 - 4ac = 3^2 - 4(2)(5) = 9 - 40 = -31$.
3
Since $\Delta < 0$, there are no real roots (the two roots are complex conjugates).

Answer:

No real roots ($\Delta = -31 < 0$).

Contest level

For what value of $c$ is the line $y = 2x + c$ tangent to the parabola $y = x^2$?

1
Tangency means the line meets the parabola in exactly one point, so $x^2 = 2x + c$ has a repeated root.
2
Rearrange: $x^2 - 2x - c = 0$, with $a = 1$, $b = -2$, constant $-c$.
3
Set $\Delta = 0$: $(-2)^2 - 4(1)(-c) = 4 + 4c = 0$, so $c = -1$.

Answer:

$c = -1$ (tangent at $x = 1$).

Olympiad / Challenge

Find all integers $n$ for which $x^2 + nx + (n + 3) = 0$ has integer roots.

1
Integer roots require $\Delta = n^2 - 4(n + 3) = n^2 - 4n - 12$ to be a perfect square, say $k^2$ with $k \ge 0$. Complete the square: $(n - 2)^2 - 16 = k^2$.
2
Rearrange to a difference of squares: $(n - 2)^2 - k^2 = 16$, i.e. $(n - 2 - k)(n - 2 + k) = 16$. The two factors have equal sums to $2(n-2)$, so they share parity; both must be even.
3
Same-parity factor pairs of $16$ (with $n - 2 - k \le n - 2 + k$) are $(2, 8), (4, 4), (-8, -2), (-4, -4)$, giving $n - 2 \in \{5, 4, -5, -4\}$, so $n \in \{7, 6, -3, -2\}$.
4
Each makes $\Delta$ a perfect square AND the roots $\frac{-n \pm k}{2}$ integers: $n=7 \Rightarrow (x+2)(x+5)$; $n=6 \Rightarrow (x+3)^2$; $n=-3 \Rightarrow x(x-3)$; $n=-2 \Rightarrow (x-1)^2$.

Answer:

$n \in \{-3, -2, 6, 7\}$.

Going Deeper

Generalization: for any polynomial the discriminant (a symmetric function of the coefficients) vanishes exactly when two roots coincide; for the cubic $x^3 + px + q$ it is $-4p^3 - 27q^2$.
Where it appears: AMC 'for how many $k$ are the roots real/rational' counting, AIME problems forcing $\Delta$ to be a perfect square for integer roots, and tangency conditions (line meets conic once $\iff \Delta = 0$).
Pitfall: $\Delta$ being a perfect square only guarantees RATIONAL roots when $a, b, c$ are integers; with non-integer coefficients that fails. Also $\Delta \ge 0$ (not $> 0$) is the condition for 'at least one real root' — do not exclude the repeated-root boundary $\Delta = 0$.

Spot the Signal

Look for problems where the key step is using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges.
You can describe the hard part as using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for quadratic discriminant, then rewrite the givens around it.
Name the relation that makes Quadratic Discriminant legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Quadratic Discriminant just because the surface looks familiar; verify the required condition first.
Applying Quadratic Discriminant because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let quadratic discriminant reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using $b^2-4ac$ to control roots, tangency, integer solutions, and parameter ranges.