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Solve quadratic inequalities with and without real roots

Find the zeros, then check the sign between them

Move every term to one side. If the quadratic has two real roots, they divide the number line into intervals with constant signs. For a positive leading coefficient, the quadratic is negative between the roots and positive outside. Strict inequalities exclude roots; non-strict inequalities include them when equality satisfies the condition.

Problem

Solve over the real numbers, giving intervals where appropriate. Example 1. $x^{2} - 36 < 0$ . Example 2. $2 x^{2} < - x + 3$ . Example 3. $3 x^{2} + 16 < 8 x$ .

Answer

Example 1:

(- 6, 6)

. Example 2:

(- 3/2, 1)

. Example 3: no real solutions.

A negative discriminant means there are no real roots, but the leading coefficient still determines the sign of the quadratic.

Step-by-step solution

1. A difference of squares

Factor $x^{2} - 36 = (x - 6) (x + 6)$ . The roots are $- 6$ and $6$ . The upward-opening quadratic is negative between its roots, so the first answer is $(- 6, 6)$ .

2. Move all terms to one side

The second inequality becomes $2 x^{2} + x - 3 < 0$ . Factor it as $(2 x + 3) (x - 1) < 0$ . Its roots are $- 3/2$ and $1$ . The product is negative between them, giving $(- 3/2, 1)$ .

3. A quadratic that never becomes negative

The third inequality becomes $3 x^{2} - 8 x + 16 < 0$ . Complete the square:

3 x^{2} - 8 x + 16 = 3 (x - 4/3)^{2} + 32/3.

The square is nonnegative and $32/3 > 0$ , so the expression is positive for every real $x$ . There are no real solutions.

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