Quadratic equations are like puzzles with x-squared terms. They pop up in many real-world situations, from physics to finance. Solving them is a key skill for the SAT Math section and beyond.

There are three main ways to crack these equations: factoring, using the quadratic formula, or completing the square. Each method has its strengths, and knowing when to use which can save you time and boost your score.

Solving quadratics by factoring

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Factoring techniques

• Quadratic equation written in standard form $ax^2 + bx + c = 0$, where a, b, and c are real numbers and $a ≠ 0$
• Zero product property states if product of two factors equals zero, at least one factor must be zero
• Factor quadratic expression by finding two numbers with product $ac$ and sum $b$
• Rewrite quadratic expression as product of two binomial factors using these numbers
• For equations in form $x^2 + bx + c = 0$ ($a = 1$), factor as $(x + m)(x + n) = 0$, where $m$ and $n$ are factors of $c$ that add up to $b$
• For leading coefficient other than 1, factor out greatest common factor (GCF) before factoring remaining quadratic expression

Solving process

• Set each factor equal to zero after factoring quadratic equation
• Solve for variable to find roots or solutions of equation
• Example: Solve $x^2 + 5x + 6 = 0$
  • Factors: $(x + 2)(x + 3) = 0$
  • Solutions: $x = -2$ or $x = -3$
• Example: Solve $2x^2 + 7x + 3 = 0$
  • Factor out GCF: $x(2x + 1) + 3(2x + 1) = 0$
  • Factor: $(2x + 1)(x + 3) = 0$
  • Solutions: $x = -1/2$ or $x = -3$

Solving quadratics with the quadratic formula

Formula and application

• Quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are coefficients of quadratic equation in standard form $ax^2 + bx + c = 0$ and $a ≠ 0$
• Solves any quadratic equation, regardless of factorability
• Simplify expression under square root first, then add and subtract square root term to find both solutions
• Example: Solve $2x^2 - 5x - 12 = 0$
  • $a = 2$, $b = -5$, $c = -12$
  • $x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-12)}}{2(2)}$
  • $x = \frac{5 \pm \sqrt{25 + 96}}{4} = \frac{5 \pm \sqrt{121}}{4} = \frac{5 \pm 11}{4}$
  • Solutions: $x = 4$ or $x = -3/2$

Discriminant analysis

• Discriminant expression $b^2 - 4ac$ under square root in quadratic formula
• Determines nature of roots
• Positive discriminant yields two distinct real roots
• Zero discriminant yields one real root (double root)
• Negative discriminant yields no real roots (two complex roots)
• Example: Analyze $x^2 + 4x + 5 = 0$
  • Discriminant $= 4^2 - 4(1)(5) = 16 - 20 = -4$
  • Negative discriminant indicates no real roots

Solving quadratics by completing the square

Completing the square process

• Method creates perfect square trinomial to solve quadratic equations
• Leading coefficient of quadratic term must be 1 (divide both sides by leading coefficient if not)
• Rewrite quadratic equation as $x^2 + bx = -c$
• Add square of half x-term coefficient to both sides: $x^2 + bx + (\frac{b}{2})^2 = (\frac{b}{2})^2 - c$
• Factor left side as perfect square: $(x + \frac{b}{2})^2 = (\frac{b}{2})^2 - c$
• Take square root of both sides and solve for $x$ to find roots

Applications and examples

• Derive quadratic formula by solving general quadratic equation $ax^2 + bx + c = 0$ using this method
• Example: Solve $x^2 + 6x - 7 = 0$
  • Rewrite: $x^2 + 6x = 7$
  • Add $(\frac{6}{2})^2 = 9$ to both sides: $x^2 + 6x + 9 = 16$
  • Factor left side: $(x + 3)^2 = 16$
  • Take square root: $x + 3 = \pm 4$
  • Solve for $x$: $x = -3 \pm 4$
  • Solutions: $x = 1$ or $x = -7$