Equations in quadratic form are a powerful tool for solving complex problems. They allow us to tackle equations that might seem impossible at first glance. By using substitution and standard quadratic solving techniques, we can break down tricky problems into manageable steps.

This topic builds on our knowledge of quadratic equations and introduces new problem-solving strategies. It shows how seemingly different types of equations can be approached with similar methods, expanding our mathematical toolkit and problem-solving abilities.

Solving Equations in Quadratic Form

Solving equations in quadratic form

• Quadratic form equations have the structure $a(expression)^2 + b(expression) + c = 0$
  • $a$, $b$, and $c$ are constants with $a \neq 0$ (e.g., $3(2x-1)^2 + 4(2x-1) - 5 = 0$)
  • The $(expression)$ is typically a binomial or a more complex term (e.g., $(3x+2)$, $(x^2-4x+1)$)
• Identify the appropriate substitution, letting $u = (expression)$ (e.g., let $u = 2x-1$)
• Rewrite the equation in terms of $u$, resulting in a quadratic equation: $au^2 + bu + c = 0$ (e.g., $3u^2 + 4u - 5 = 0$)
• Solve the quadratic equation using standard techniques
  • Factoring (e.g., $(3u-5)(u+1) = 0$, so $u = \frac{5}{3}$ or $u = -1$)
  • Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  • Completing the square
• Substitute the original $(expression)$ back in for $u$ to find the solution(s) to the original equation (e.g., $2x-1 = \frac{5}{3}$ or $2x-1 = -1$, so $x = \frac{11}{6}$ or $x = 0$)
• This process often involves variable substitution, a key technique in solving polynomial equations

Substitution for rational exponents

• Equations with rational exponents can often be transformed into quadratic form using substitution
• For equations with square roots, let $u = \sqrt{expression}$ (e.g., $\sqrt{2x+1} = 3$ becomes $u = \sqrt{2x+1}$)
• For equations with cube roots, let $u = \sqrt[3]{expression}$ (e.g., $\sqrt[3]{x^2-1} = 2$ becomes $u = \sqrt[3]{x^2-1}$)
• For equations with fractional exponents, let $u = (expression)^{\frac{1}{n}}$, where $n$ is the denominator of the exponent (e.g., $(3x-2)^{\frac{2}{3}} = 4$ becomes $u = (3x-2)^{\frac{1}{3}}$)
• After making the substitution, rewrite the equation in terms of $u$
  • Raise both sides of the equation to the appropriate power to eliminate the rational exponent (e.g., $u^2 = 9$, $u^3 = 8$, $u^3 = 64$)
  • Simplify the equation to obtain a quadratic equation in terms of $u$ (e.g., $u^2 = 9$ becomes $u^2 - 9 = 0$)
• Solve the resulting quadratic equation using standard techniques
• Substitute the original $(expression)$ back in for $u$ to find the solution(s) to the original equation

Evaluating quadratic form solutions

• After solving a quadratic form equation, evaluate the solutions to determine their validity
• Extraneous roots may arise when the original equation involves rational expressions or fractions or the equation is transformed by raising both sides to an even power
• Substitute each solution back into the original equation and simplify to see if it results in a true statement
  • If a solution does not satisfy the original equation, it is an extraneous root and should be discarded
• When an equation is transformed by raising both sides to an even power, consider the possibility of introducing new solutions that satisfy the transformed equation but not the original
  • Check the solutions in the original equation to identify and eliminate any extraneous roots (e.g., if $\sqrt{x+1} = x-3$, squaring both sides gives $x+1 = (x-3)^2$, which has solutions $x = 0$ and $x = 7$, but only $x = 7$ satisfies the original equation)

Advanced Techniques and Considerations

• Some polynomial equations can be solved using quadratic form techniques through clever algebraic manipulation
• Even functions, which are symmetric about the y-axis, often lead to equations in quadratic form when solving for their roots
• The process of solving equations in quadratic form often requires a combination of algebraic skills and recognition of patterns to simplify complex expressions