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GED Math: Quadratic Equations Concepts

Importance ★★★☆☆ · 15% (estimate)

Overview

This chapter covers quadratic equations, which have an x² term. You will solve them by taking square roots, by factoring, and by using the quadratic formula, and you will write quadratic equations for situations involving area, consecutive numbers, and objects thrown in the air.

Before factoring, move every term to one side so the equation equals zero. If a product equals zero, at least one factor must be zero, so each factor is set equal to zero and solved. An equation such as x² = 49 has two answers, 7 and −7, because both square to 49.

When an equation does not factor easily, use the quadratic formula from the formula sheet. Identify a, b, and c from the standard form ax² + bx + c = 0 and substitute carefully, especially when b or c is negative. In real situations, throw away answers that make no sense, such as a negative length.

Key concepts

  • Standard form

    A quadratic equation in standard form is ax² + bx + c = 0 (the formula sheet writes the function as y = ax² + bx + c).

    Memory tipMove all terms to one side before factoring or using the formula.

    Source: GED Testing Service (Mathematics Formula Sheet) · as of 2026

  • Zero product property

    If (x − 3)(x + 5) = 0, then x − 3 = 0 or x + 5 = 0, so x = 3 or x = −5.

    Memory tipThe answers have the opposite signs of the numbers in the factors.

  • Square root method

    When the equation is x² = k, x = √k or x = −√k. Get x² alone first.

    Memory tipDo not forget the negative answer.

  • Quadratic formula

    x = [−b ± √(b² − 4ac)] / 2a. The part under the root, b² − 4ac, is computed first.

    Memory tipIf b is negative, −b is positive.

    Source: GED Testing Service (Mathematics Formula Sheet) · as of 2026

  • Quadratics in context

    Area problems (w × (w + 4) = 96), consecutive numbers (n(n + 1) = 132), and heights of thrown objects (h = −16t² + 64t) lead to quadratic equations.

    Memory tipKeep only answers that make sense, such as positive lengths and times.

Often tested

  • Set the equation equal to zero before factoring.
  • Check each answer by substituting it back into the original equation.
  • For a thrown object, "lands" or "hits the ground" means height = 0.
  • Divide every part of the numerator by 2a when simplifying the quadratic formula.

3 sample questions

Verified practice questions from this unit.

Practice questionReviewedAlgebraic Problem Solving › Quadratic Equations★★★☆☆

Question 1. Solve: x² = 49

  1. ① x = 2,401
  2. ② x = 7 or x = −7
  3. ③ x = 24.5 or x = −24.5
  4. ④ x = 7 only
▼ Show answer and explanation▲ Hide answer and explanation

Answer: ② x = 7 or x = −7

Key point: x² = k has two answers: positive and negative √k

  1. Take the square root of both sides.
  2. Both 7 × 7 and (−7) × (−7) equal 49.
  3. So x = 7 or x = −7.

Check: 7² = 49 and (−7)² = 49.

Wrong choices

  • x = 7 only: x = 7 only misses the negative answer; (−7)² is also 49.
  • x = 24.5 or x = −24.5: x = 24.5 or x = −24.5 divides 49 by 2 instead of taking the square root.
  • x = 2,401: x = 2,401 squares 49 instead of taking its square root.
Practice questionReviewedAlgebraic Problem Solving › Quadratic Equations★★★☆☆

Question 2. Solve: (x − 3)(x + 5) = 0

  1. ① x = −3 or x = 5
  2. ② x = 3 or x = 5
  3. ③ x = −3 or x = −5
  4. ④ x = 3 or x = −5
▼ Show answer and explanation▲ Hide answer and explanation

Answer: ④ x = 3 or x = −5

Key point: If a product is 0, set each factor equal to 0

  1. Set each factor equal to 0.
  2. x − 3 = 0 gives x = 3.
  3. x + 5 = 0 gives x = −5.

Check: (3 − 3)(3 + 5) = 0 × 8 = 0, and (−5 − 3)(−5 + 5) = −8 × 0 = 0.

Wrong choices

  • x = −3 or x = 5: x = −3 or x = 5 takes the numbers with the signs shown in the factors; the signs must be reversed.
  • x = 3 or x = 5: x = 3 or x = 5 gets the sign of the second answer wrong; 5 + 5 is not 0.
  • x = −3 or x = −5: x = −3 or x = −5 gets the sign of the first answer wrong; −3 − 3 is not 0.
Practice questionReviewedAlgebraic Problem Solving › Quadratic Equations★★★☆☆

Question 3. Solve: x² − 7x + 12 = 0

  1. ① x = 2 or x = 6
  2. ② x = 1 or x = 12
  3. ③ x = −3 or x = −4
  4. ④ x = 3 or x = 4
▼ Show answer and explanation▲ Hide answer and explanation

Answer: ④ x = 3 or x = 4

Key point: Factor, then set each factor equal to zero

  1. Find two numbers that multiply to 12 and add to −7: −3 and −4.
  2. Factor: (x − 3)(x − 4) = 0.
  3. x = 3 or x = 4.

Check: 3² − 7(3) + 12 = 9 − 21 + 12 = 0, and 4² − 7(4) + 12 = 16 − 28 + 12 = 0.

Wrong choices

  • x = −3 or x = −4: x = −3 or x = −4 uses the numbers from the factors without reversing the signs; (−3)² − 7(−3) + 12 = 42.
  • x = 2 or x = 6: x = 2 or x = 6 multiply to 12 but add to 8, not 7; 2² − 14 + 12 = 2.
  • x = 1 or x = 12: x = 1 or x = 12 multiply to 12 but add to 13; 1 − 7 + 12 = 6.
Practice 5 questions from this unit →

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