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3.3 Quadratic Equations & Quadratic Inequalities

Key Takeaways

  • Quadratic equations ax^2 + bx + c = 0 (a != 0) can be solved by factoring, using the square root property, completing the square, or applying the quadratic formula x = (-b +- sqrt(b^2 - 4ac))/(2a).

  • The discriminant D = b^2 - 4ac determines the number and nature of roots: D > 0 gives two real roots, D = 0 gives one repeated real root (double root), and D < 0 gives two complex conjugate roots.

  • Completing the square transforms ax^2 + bx + c = 0 into vertex form a(x - h)^2 + k = 0, forming the algebraic basis of the quadratic formula and parabola coordinate shifts.

  • Solving quadratic inequalities requires finding critical boundary points (roots) and evaluating open test intervals using a sign chart.

Last updated: August 2026

3.3 Quadratic Equations & Quadratic Inequalities

A quadratic equation is a second-degree polynomial equation in one variable. The standard form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

where a,b,a, b, and cc are real numbers with a≠0a \neq 0. The exponent 22 makes this a non-linear equation whose graphs form parabolas.


1. The Four Core Algebraic Solving Methods

Depending on the algebraic structure of a quadratic equation, four distinct methods can be employed.

Method 1: Factoring and Zero-Product Property

If a quadratic expression ax2+bx+cax^2 + bx + c can be factored into linear binomials (px+q)(rx+s)=0(px + q)(rx + s) = 0, the Zero-Product Property states that at least one factor must equal zero:

px+q=0orrx+s=0px + q = 0 \quad \text{or} \quad rx + s = 0

Worked Example 1: Factoring

Solve 3x2−7x−6=03x^2 - 7x - 6 = 0.

  1. Find two numbers that multiply to a⋅c=3(−6)=−18a \cdot c = 3(-6) = -18 and add to b=−7b = -7. Those numbers are −9-9 and +2+2.
  2. Rewrite the middle term: 3x2−9x+2x−6=03x^2 - 9x + 2x - 6 = 0.
  3. Factor by grouping: 3x(x−3)+2(x−3)=(3x+2)(x−3)=03x(x - 3) + 2(x - 3) = (3x + 2)(x - 3) = 0.
  4. Set factors to zero: 3x+2=0  ⟹  x=−233x + 2 = 0 \implies x = -\frac{2}{3} x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3

Solution set: {−23,3}\left\{-\frac{2}{3}, 3\right\}.

Method 2: The Square Root Property

If an equation can be written in the form u2=ku^2 = k (where uu is an algebraic expression and k≥0k \ge 0), then:

u=±ku = \pm \sqrt{k}

Worked Example 2: Square Root Property

Solve (2x−5)2=12(2x - 5)^2 = 12.

  1. Apply square root to both sides: 2x−5=±12=±232x - 5 = \pm \sqrt{12} = \pm 2\sqrt{3}
  2. Isolate xx: 2x=5±23  ⟹  x=5±2322x = 5 \pm 2\sqrt{3} \implies x = \frac{5 \pm 2\sqrt{3}}{2}

Method 3: Completing the Square

Completing the square transforms any quadratic polynomial into a perfect square trinomial plus a constant.

Algorithm:

  1. Rewrite ax2+bx+c=0ax^2 + bx + c = 0 as x2+bax=−cax^2 + \frac{b}{a}x = -\frac{c}{a} (divide by aa).
  2. Add (b2a)2\left(\frac{b}{2a}\right)^2 to both sides.
  3. Factor the left side as (x+b2a)2\left(x + \frac{b}{2a}\right)^2.
  4. Solve using the Square Root Property.

Worked Example 3: Completing the Square

Solve 2x2+8x−10=02x^2 + 8x - 10 = 0.

  1. Divide by 2: x2+4x−5=0  ⟹  x2+4x=5x^2 + 4x - 5 = 0 \implies x^2 + 4x = 5.
  2. Half of coefficient 4 is 2; square it (22=42^2 = 4). Add 4 to both sides: x2+4x+4=5+4  ⟹  (x+2)2=9x^2 + 4x + 4 = 5 + 4 \implies (x + 2)^2 = 9
  3. Take square roots: x+2=±3x + 2 = \pm 3.
  4. x=−2+3=1x = -2 + 3 = 1 or x=−2−3=−5x = -2 - 3 = -5. Solution set: {−5,1}\{-5, 1\}.

Method 4: The Quadratic Formula

Derived directly by completing the square on the general equation ax2+bx+c=0ax^2 + bx + c = 0, the Quadratic Formula solves any quadratic equation:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

2. Discriminant Analysis (D=b2−4acD = b^2 - 4ac)

The expression under the radical in the quadratic formula, D=b2−4acD = b^2 - 4ac, is called the discriminant. It dictates the number and numerical character of the roots without fully completing the radical arithmetic.

Discriminant Value (D=b2−4acD = b^2 - 4ac)Nature of RootsGraph Intercepts (y=ax2+bx+cy = ax^2 + bx + c)
D>0D > 0, Perfect Square2 Distinct Rational Real Roots2 distinct xx-intercepts
D>0D > 0, Non-Perfect Square2 Distinct Irrational Real Roots2 distinct xx-intercepts
D=0D = 01 Repeated Real Root (Double Root)1 xx-intercept (vertex touches xx-axis)
D<0D < 02 Complex Conjugate Non-Real Roots (u±viu \pm vi)No xx-intercepts (parabola floats above/below axis)

Worked Example 4: Quadratic Formula with Complex Conjugate Roots

Problem: Solve 2x2−4x+5=02x^2 - 4x + 5 = 0.

Step-by-Step Solution:

  1. Identify coefficients: a=2,b=−4,c=5a = 2, b = -4, c = 5.
  2. Compute discriminant: D=b2−4ac=(−4)2−4(2)(5)=16−40=−24D = b^2 - 4ac = (-4)^2 - 4(2)(5) = 16 - 40 = -24
  3. Since D=−24<0D = -24 < 0, the roots are complex conjugates involving i=−1i = \sqrt{-1}: x=−(−4)±−242(2)=4±i244=4±2i64=1±62ix = \frac{-(-4) \pm \sqrt{-24}}{2(2)} = \frac{4 \pm i\sqrt{24}}{4} = \frac{4 \pm 2i\sqrt{6}}{4} = 1 \pm \frac{\sqrt{6}}{2}i

3. Solving Quadratic Inequalities & Sign Charts

A quadratic inequality can be written in forms such as ax2+bx+c>0ax^2 + bx + c > 0, ax2+bx+c≤0ax^2 + bx + c \le 0, etc.

Step-by-Step Sign Chart Algorithm

  1. Standard Form: Move all terms to the left side so that zero appears on the right side.
  2. Find Critical Numbers: Solve the boundary equation ax2+bx+c=0ax^2 + bx + c = 0 to find real roots.
  3. Partition Test Intervals: Plot critical numbers on a real number line to divide it into open test intervals.
  4. Evaluate Interval Signs: Select a test value xtx_t within each open interval and substitute it into the quadratic polynomial to record its sign (++ or −-).
  5. State Solution: Select intervals that satisfy the inequality operator (>> or ≥\ge requires ++; << or ≤\le requires −-). Include endpoints for ≤,≥\le, \ge unless restricted.

Worked Example 5: Quadratic Inequality

Problem: Solve x2−2x−8≥0x^2 - 2x - 8 \ge 0.

Step-by-Step Solution:

  1. Factor: (x−4)(x+2)≥0(x - 4)(x + 2) \ge 0.
  2. Critical boundary points: x=−2x = -2 and x=4x = 4.
  3. Test intervals: (−∞,−2)(-\infty, -2), (−2,4)(-2, 4), and (4,∞)(4, \infty).
  4. Test point evaluation:
    • Interval (−∞,−2)(-\infty, -2), test x=−3x = -3: (−3−4)(−3+2)=(−7)(−1)=+7>0(-3 - 4)(-3 + 2) = (-7)(-1) = +7 > 0 (+)
    • Interval (−2,4)(-2, 4), test x=0x = 0: (0−4)(0+2)=(−4)(2)=−8<0(0 - 4)(0 + 2) = (-4)(2) = -8 < 0 (-)
    • Interval (4,∞)(4, \infty), test x=5x = 5: (5−4)(5+2)=(1)(7)=+7>0(5 - 4)(5 + 2) = (1)(7) = +7 > 0 (+)
  5. Inequality calls for ≥0\ge 0 (positive or zero). Endpoints −2-2 and 44 are included. Solution set: (−∞,−2]∪[4,∞)(-\infty, -2] \cup [4, \infty).

Worked Example 6: Strict Quadratic Inequality

Problem: Solve 2x2−5x−3<02x^2 - 5x - 3 < 0.

Step-by-Step Solution:

  1. Factor: (2x+1)(x−3)<0(2x + 1)(x - 3) < 0.
  2. Critical points: x=−12x = -\frac{1}{2} and x=3x = 3.
  3. Test intervals: (−∞,−12)\left(-\infty, -\frac{1}{2}\right), (−12,3)\left(-\frac{1}{2}, 3\right), and (3,∞)(3, \infty).
  4. Evaluating test points reveals negative output only on (−12,3)\left(-\frac{1}{2}, 3\right).
  5. Since the inequality is strictly <<, endpoints are excluded. Solution set: (−12,3)\left(-\frac{1}{2}, 3\right).
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Quadratic Solving Method Selection Architecture
Test Your Knowledge

What are the solutions to the quadratic equation x2−4x+13=0x^2 - 4x + 13 = 0?

A

x = 2 +- 3i

B

x = -2 +- 3i

C

x = 2 +- 6i

D

x = 4 +- 6i

Test Your Knowledge

What is the solution set of the quadratic inequality 2x2−5x−3<02x^2 - 5x - 3 < 0?

A

(-inf, -1/2) U (3, inf)

B

(-1/2, 3)

C

[-1/2, 3]

D

(-inf, -3) U (1/2, inf)

Test Your Knowledge

For what value of kk does the quadratic equation x2−6x+k=0x^2 - 6x + k = 0 have exactly one real solution?

A

k = 0

B

k = 3

C

k = 9

D

k = 36

Test Your Knowledge

What are the real solutions to the equation (2x−1)2=25(2x - 1)^2 = 25?

A

x = 3 and x = -2

B

x = 3 only

C

x = 13 and x = -12

D

x = 6 and x = -4

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