Nature of Roots vs Sign of a Quadratic: Choose the Right Test

You calculate b² - 4ac, get a positive answer, and write: “The quadratic is positive.”

It feels reasonable. There is a positive number in your working, after all. But that number describes the roots of an equation, not the value of the expression at every x.

For example, x² - 8x + 12 has a positive discriminant and two positive roots. Yet at x = 4, its value is -4.

This is the small distinction behind a surprisingly large number of mistakes. Once you separate the questions, the rules become much easier to choose.

Throughout this guide, coefficients and x are real unless stated otherwise. We write f(x) = ax² + bx + c, with a ≠ 0, and D = b² - 4ac. The notation sqrt means the nonnegative square root.

First, identify which question you are answering

Look at the exact words before choosing a formula.

Wording in the question	What you need to establish	First tool
Two distinct real roots	Two different solutions of f(x) = 0	D > 0
Equal real roots	One repeated real solution	D = 0
No real roots	No real solution of f(x) = 0	D < 0
Both roots positive	Both solutions are greater than zero	Reality, sum and product
f(x) > 0 for every real x	Every output is strictly positive	a > 0 and D < 0
Solve f(x) > 0	Find the inputs giving positive outputs	Roots and sign analysis

“The roots are positive” is a statement about x-values. “The quadratic is positive” is a statement about y-values, or outputs.

On a graph, positive roots are intercepts to the right of the y-axis. Positive values are points above the x-axis. These are different directions on the same picture.

One expression, three answers

Take f(x) = x² - 8x + 12 = (x - 2)(x - 6).

• Nature of roots: D = 64 - 48 = 16 > 0, so there are two distinct real roots.
• Signs of roots: the roots 2 and 6 are both positive.
• Sign of the expression: f(x) is negative for 2 < x < 6 and positive for x < 2 or x > 6. At 2 and 6, it is zero.

So neither “D is positive” nor “both roots are positive” means “the expression is always positive.”

What the discriminant actually tells you

The quadratic formula is:

x = [-b ± sqrt(D)] / (2a).

If D > 0, the square root is positive, so adding it and subtracting it produce two different real answers. If D = 0, both choices produce the same real answer. If D < 0, there is no real square root, so there are no real roots.

This classification is explained in NCERT’s Quadratic Equations chapter, Section 4.4.

Discriminant	Roots over the real numbers	Graph and x-axis
D > 0	Two distinct real roots	Crosses at two points
D = 0	One distinct real root, counted twice	Touches at the vertex
D < 0	No real roots	Does not meet the axis

For real coefficients, D < 0 still gives two non-real complex conjugate roots when complex numbers are allowed. Write “no real roots”, rather than “no roots”.

Example 1: classify without solving

Consider 5x² + 4x + 3 = 0.

D = 4² - 4(5)(3) = 16 - 60 = -44.

Answer: two non-real complex conjugate roots; no real roots.

That completes a question asking only for the nature of the roots. Finding the exact complex roots would be extra work unless requested.

Does D > 0 also tell you whether roots are rational?

Not by itself. For a quadratic with rational coefficients, the roots are rational when D is the square of a rational number. With integer coefficients, this becomes the familiar nonnegative perfect-square test.

For example, x² - 8x + 12 = 0 has D = 16 and rational roots 2 and 6. In contrast, x² - 8x + 11 = 0 has D = 20 and irrational roots 4 ± sqrt(5).

Both equations have two distinct real roots. The difference is rationality.

How to find the sign of a quadratic expression

The graph y = f(x) opens upward when a > 0 and downward when a < 0. Its roots, when real, mark where the graph reaches zero.

Case 1: two distinct real roots

Let the roots be r and s, with r < s. Then:

f(x) = a(x - r)(x - s).

To the left of r, both bracketed factors are negative. Between r and s, one is positive and the other negative. To the right of s, both are positive.

Multiplying by a gives this complete pattern:

Location of x	a > 0	a < 0
x < r	Positive	Negative
r < x < s	Negative	Positive
x > s	Positive	Negative
x = r or x = s	Zero	Zero

A useful phrase is same sign as a outside the roots, opposite sign between them. It works for positive and negative leading coefficients.

Example 2: a negative leading coefficient

Solve -3x² + 9x + 12 > 0.

Factorise:

f(x) = -3(x - 4)(x + 1).

The roots are -1 and 4. Since a = -3, the expression is positive between the roots.

Answer: -1 < x < 4.

Check x = 0: f(0) = 12 > 0. Check x = 5: f(5) = -18 < 0. Both agree with the sign pattern.

If the question used ≥ instead of >, the answer would be -1 ≤ x ≤ 4 because the endpoints make the expression zero.

Case 2: a repeated real root

When D = 0, write f(x) = a(x - r)².

A square is positive away from r and zero at r. Therefore the expression has the sign of a everywhere except at its repeated root, where it is zero. The sign does not change across a repeated root.

For f(x) = 2x² - 12x + 18 = 2(x - 3)²:

• f(x) ≥ 0 for every real x.
• f(x) > 0 for every real x except 3.
• f(x) ≤ 0 only when x = 3.
• f(x) < 0 has no real solution.

Calling this expression “always positive” would lose the equality case. It is always nonnegative.

Case 3: no real roots

When D < 0, the graph never meets the x-axis. Its sign is the sign of a for every real x.

Thus 5x² + 4x + 3, whose discriminant is -44, is always positive because a = 5 > 0. The expression -5x² - 4x - 3 is always negative. Its discriminant is also -44, but its leading coefficient is negative.

This pair shows why D < 0 alone does not mean positive.

For more practice with interval answers and endpoints, see our quadratic inequalities and sign-chart guide. The graphical connection between inequalities and regions above or below the axis is also covered in OpenStax’s quadratic inequalities lesson.

Why a and D work together: complete the square

There is one identity that connects all these rules:

f(x) = a[x + b/(2a)]² - D/(4a).

You can verify it by expanding the square. The x² and x terms become ax² + bx, while the constant becomes (b² - D)/(4a) = c.

The square becomes zero at x = -b/(2a). This is the vertex input, and the vertex value is:

f(-b/(2a)) = -D/(4a).

When a > 0, this value is the minimum. When a < 0, it is the maximum. The interpretation of the vertex as a minimum or maximum is explained in OpenStax’s quadratic functions chapter.

For an upward-opening graph to stay strictly above zero, its minimum must be positive. With a > 0, the condition -D/(4a) > 0 becomes D < 0.

For a downward-opening graph to stay strictly below zero, its maximum must be negative. With a < 0, the condition -D/(4a) < 0 again becomes D < 0.

That gives four useful tests, all assuming a ≠ 0:

Required for every real x	Necessary and sufficient conditions
f(x) > 0	a > 0 and D < 0
f(x) ≥ 0	a > 0 and D ≤ 0
f(x) < 0	a < 0 and D < 0
f(x) ≤ 0	a < 0 and D ≤ 0

Parameter questions: translate the words before calculating

A letter such as k or m usually controls the coefficients. Write the required conditions first, then solve them together.

Example 3: the same family, three different questions

Let f(x) = x² - 6x + k, where k is real.

Here a = 1 and D = 36 - 4k.

Question A: when does f(x) = 0 have real roots?

We need D ≥ 0:

36 - 4k ≥ 0, so k ≤ 9.

Question B: when is f(x) positive for every real x?

We need a > 0 and D < 0. The first condition already holds:

36 - 4k < 0, so k > 9.

Question C: when is f(x) nonnegative for every real x?

Allow the graph to touch zero, so D ≤ 0:

36 - 4k ≤ 0, so k ≥ 9.

Check the boundary directly. At k = 9, f(x) = (x - 3)². It is nonnegative everywhere, but f(3) = 0, so it is not strictly positive everywhere.

Example 4: when the coefficient of x² can disappear

Find all real m for which:

(m - 2)x² + 4x + m + 1 > 0 for every real x.

First check m = 2. The expression becomes 4x + 3, which is negative for sufficiently negative x. So m = 2 does not work.

For m ≠ 2, it is quadratic. We need:

1. a = m - 2 > 0, giving m > 2.
2. D = 16 - 4(m - 2)(m + 1) < 0.

Simplify the discriminant:

D = -4(m² - m - 6) = -4(m - 3)(m + 2).

Thus D < 0 means (m - 3)(m + 2) > 0, giving m < -2 or m > 3.

Intersect this with m > 2:

Answer: m > 3.

At m = 3, the expression becomes (x + 2)², so strict positivity fails at x = -2. For m < -2, D is negative but a is negative too: the expression is always negative, which is the opposite of the requirement.

Example 5: a constant case that must be included

Find all real k for which:

kx² + 4 > 0 for every real x.

For k > 0, kx² ≥ 0, so the whole expression is at least 4. For k = 0, the expression is the constant 4, so it still works. For k < 0, sufficiently large values of |x| make kx² + 4 negative.

Answer: k ≥ 0.

If you had applied only the quadratic rule a > 0 and D < 0, you would have obtained k > 0 and missed the valid constant case.

Positive roots are a separate problem

Suppose the roots are r and s. Expanding a(x - r)(x - s) and comparing coefficients gives:

r + s = -b/a, and rs = c/a.

If the roots are real and their product is positive, they have the same nonzero sign. A positive sum then makes both positive; a negative sum makes both negative.

Therefore:

• Both roots positive: D ≥ 0, -b/a > 0, and c/a > 0.
• Both roots negative: D ≥ 0, -b/a < 0, and c/a > 0.
• Roots of opposite signs: c/a < 0. For real coefficients this already guarantees D > 0, since ac < 0.

If the question requires two distinct positive or negative roots, replace D ≥ 0 with D > 0. If c = 0, zero is a root and is neither positive nor negative.

The sum-product identities are listed in SATHEE’s quadratic equations formula reference. The conditions above follow by checking the possible signs of two real numbers.

Example 6: both roots positive, allowing equality

Find k such that x² - 6x + k = 0 has two positive roots, allowing a repeated root.

The conditions are:

• D = 36 - 4k ≥ 0, so k ≤ 9.
• Sum = 6 > 0, which is already satisfied.
• Product = k > 0.

Answer: 0 < k ≤ 9.

For two distinct positive roots, the answer becomes 0 < k < 9.

Compare this with Example 3. The expression is positive for every real x when k > 9, but its equation has positive real roots when 0 < k ≤ 9. Confusing the wording produces entirely different parameter ranges.

Why the reality check cannot be skipped

For x² - 2x + 5 = 0, the sum is 2 and the product is 5. Both are positive. But D = 4 - 20 = -16, so the roots are not real and cannot be described as positive real numbers.

A positive sum and product are not enough on their own.

What changes when x has a restricted domain?

The all-real tests answer a strong question: does the condition hold at every real input?

A restriction such as 0 ≤ x ≤ 1 can change the answer. For example:

f(x) = x² - 8x + 12.

This expression is not positive for all real x. However, on [0, 1], both x - 2 and x - 6 are negative, so their product is positive throughout the interval.

For checking strict positivity on a closed interval [p, q], find the minimum on that interval:

• If a > 0 and the vertex lies inside the interval, check the vertex value.
• If a > 0 and the vertex lies outside the interval, the minimum is at an endpoint.
• If a < 0, the minimum on the closed interval is at an endpoint, so check both endpoints.

For instance, f(x) = x² - 4x + 3 has f(0) = 3 and f(4) = 3, both positive. Yet the vertex x = 2 lies inside [0, 4], and f(2) = -1. Checking only the two endpoints would miss the negative middle.

Also distinguish “for every x” from “for at least one x”. The expression x² - 1 is positive at x = 2, but not at x = 0. One successful substitution proves existence; it does not prove an all-real claim.

A short answer-writing routine

Before doing arithmetic, underline the words that tell you the task: real, distinct, positive roots, positive expression, every x, or a given interval.

Then write your conditions. For example:

“For strict positivity for all real x, a > 0 and D < 0.”

Substitute the coefficients, solve each condition, and take their intersection. Check parameter values that make a zero, and test any equality boundary that seems uncertain.

Finally, write a sentence that answers the actual question. “D = 25” is working. “The equation has two distinct real roots” is the conclusion.

Try these before reading the answers

1. Determine the nature of the roots of 4x² + 4x + 5 = 0.
2. Solve -2x² + 10x - 12 ≥ 0.
3. Find k for which x² + 8x + k is strictly positive for every real x.
4. Find k for which x² + 8x + k is nonnegative for every real x.
5. Find k for which x² - 10x + k = 0 has two distinct positive roots.
6. Find t for which (t - 1)x² + 5 is positive for every real x, including any nonquadratic case.
7. Is x² - 10x + 21 positive throughout [0, 2], even though its discriminant is positive?
8. Solve (x + 4)² ≤ 0 and then solve (x + 4)² < 0.

Answers and the reason behind each one

1. No real roots. D = 16 - 80 = -64. Over the complex numbers there are two non-real conjugate roots.

2. 2 ≤ x ≤ 3. The expression is -2(x - 2)(x - 3). A negative leading coefficient gives positive values between the roots, and ≥ includes both roots.

3. k > 16. Here a = 1 > 0 and D = 64 - 4k must be negative. Equivalently, write the expression as (x + 4)² + k - 16 and require its minimum to be positive.

4. k ≥ 16. Equality is now allowed. At k = 16, the minimum is zero at x = -4.

5. 0 < k < 25. The positive sum is 10, the product requires k > 0, and two distinct real roots require 100 - 4k > 0.

6. t ≥ 1. For t > 1 the expression is at least 5. At t = 1 it is the constant 5. For t < 1 it eventually becomes negative as |x| grows.

7. Yes. Factorise as (x - 3)(x - 7). Throughout [0, 2], both factors are negative, so their product is positive. Real roots outside the interval do not prevent positivity inside it.

8. The first has only x = -4; the second has no real solution. A square can equal zero but cannot be negative.

Frequently asked questions

Does a positive discriminant mean the quadratic is positive?

No. D > 0 means the equation has two distinct real roots. The expression changes sign at those roots. Its leading coefficient determines whether it is positive outside the roots or between them.

Can a quadratic have two positive roots but take negative values?

Yes. For example, x² - 8x + 12 has positive roots 2 and 6, but its value is negative between them. Positive roots describe inputs where the output is zero, not the signs of all outputs.

What conditions make a quadratic positive for all real x?

For real coefficients and a ≠ 0, you need a > 0 and D < 0. The graph must open upward and remain entirely above the x-axis.

Why is D = 0 excluded from strict positivity?

Because D = 0 gives a repeated real root. At that input, the expression equals zero. Zero satisfies a nonnegative condition but not a strictly positive one.

What if the discriminant is negative and a is negative?

The expression is negative for every real x. The graph opens downward and has no x-intercepts, so it stays below the x-axis.

How do I check whether both roots are positive without finding them?

Check D ≥ 0, -b/a > 0, and c/a > 0. Use D > 0 if the roots must be distinct. Reality, positive sum, and positive product must all hold together.

Does the sign change at a repeated root?

No. A quadratic with repeated root r is a(x - r)². It has the sign of a on both sides of r and equals zero at r.

What should I do if a parameter makes the x² coefficient zero?

Substitute that parameter value into the original expression and examine the resulting linear or constant case separately. It may or may not satisfy the question, and the quadratic discriminant rule cannot decide it.

Can I use the all-real positivity test on a restricted interval?

It is sufficient but not necessary. A quadratic may be positive on the requested interval while negative elsewhere. Check the sign on that interval or locate its minimum there.

What is the fastest way to avoid mixing up these questions?

Ask what must be positive: the discriminant, the roots, or the expression’s outputs. Then identify whether the statement concerns some inputs, all real inputs, or only a specified interval. Those two decisions tell you which test belongs in your answer.