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Quadratic Inequalities

Solve quadratic inequalities using roots, sign charts, and the discriminant. Four worked examples cover interval endpoints, repeated roots, and coefficients.

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Quadratic Inequalities

Solving a quadratic inequality means finding every real value of $x$ for which a quadratic expression has the required sign. The roots tell you where its graph meets the horizontal axis; the leading coefficient tells you which way the parabola opens.

1. Put the inequality in a useful form

Move all terms to one side so that the other side is zero:

If $a<0$, you can multiply both sides of the inequality by $-1$ to make the leading coefficient positive. Reverse the inequality sign when you do this. This step is convenient, not compulsory: you can also work directly with a downward-opening parabola.

Next, calculate the discriminant $\Delta=b^2-4ac$. It tells you whether there are two distinct real roots, one repeated root, or no real roots.

2. Two distinct roots

Suppose $a>0$ and $\Delta>0$, with roots $x_1<x_2$. Then

Check the signs of the factors in each interval:

For $f(x)>0$, choose the two outside intervals. For $f(x)<0$, choose the interval between the roots. Include the roots for $\ge0$ or $\le0$, since $f(x)=0$ there. Strict inequalities exclude them.

If $a<0$ and you have not multiplied the inequality by $-1$, all three signs in the last column are reversed.

3. A repeated root or no real roots

The following statements assume $a>0$.

• If $\Delta=0$, write $f(x)=a(x-r)^2$, where $r=-b/(2a)$. It is positive except at $r$, where it is zero. Thus $f(x)>0$ has solution $\mathbb{R}\setminus\{r\}$, while $f(x)<0$ has no solution. The solutions to $f(x)\ge0$ and $f(x)\le0$ are $\mathbb{R}$ and $\{r\}$, respectively.
• If $\Delta<0$, the upward-opening parabola never reaches the horizontal axis, so $f(x)>0$ for every real $x$. The inequalities $f(x)>0$ and $f(x)\ge0$ hold for all real numbers; $f(x)<0$ and $f(x)\le0$ have no solution.

A negative discriminant means the equation has no real roots. It does not, by itself, mean the inequality has no solution.

4. Worked examples

Example 1: Two outside intervals

Solve $x^2-5x+6>0$. Factor the expression:

The leading coefficient is positive and the roots are $2$ and $3$. The expression is positive outside the roots. Equality is not allowed, so the solution is

Example 2: A negative leading coefficient

Solve $-2x^2+4x-3\ge0$. Multiplying by $-1$ gives

Here $\Delta=16-24=-8$. The new quadratic is always positive, so it cannot be less than or equal to zero. The solution set is $\varnothing$.

Completing the square gives the same conclusion: $2x^2-4x+3=2(x-1)^2+1>0$.

Example 3: One included value

Solve $x^2-4x+4\le0$. Since

the expression is nonnegative and equals zero only at $x=2$. The solution set is $\{2\}$.

Example 4: Recovering the coefficients

The solution set of $x^2+ax+b<0$ is $(1,3)$. Find $a$ and $b$.

The leading coefficient is $1$, so the endpoints are the two roots:

Comparing coefficients gives $a=-4$ and $b=3$.

Check the endpoints and the original conditions

Substitute a value from each proposed interval into the original inequality. Check whether the inequality is strict, and keep any restrictions introduced by an earlier fraction or logarithm. If a parameter can make $a=0$, handle that linear or constant case separately.

Further reading

• OpenStax Intermediate Algebra: Solving quadratic inequalities

Content reviewed: 9 September 2026.