Solving+Quadratic+Equations-Target+B-Appropriate+Quadratics-Guided+Learning

Solving Quadratic Equations Target B: Determine an appropriate method for solving a quadratic equation.
Ed thinks that this quadratic equation is best solved with using square roots. Barry thinks that the equation is best solved with factoring. Whom do you agree with and why? ||
 * [[image:Discuss this.png width="208" height="116"]] || [[image:SQE B GL 1a.png]]

In Target A, there were four approaches to solving quadratic equations. For this target, we will investigate the most efficient method for solving based on a given equation. The criteria listed below can help in determining which of the four methods would be most appropriate to use to solve a quadratic equation.


 * Solving By Factoring:**
 * If you can factor out a GCF to eliminate the x 2 or quadratic term
 * If the leading coefficient is 1 or a prime number and you can identify as a difference of two squares, a perfect square trinomial, or can easily factor using guess and check


 * Solving Using Square Roots**
 * If there is an “x 2 ” or quadratic term but no “x” or linear term
 * If the equation has an expression that is of the form (x ± a) 2


 * Solving Using Completing the Square**
 * If you cannot factor the polynomial
 * If the leading coefficient is 1 (or the leading coefficient is 1 after factoring out a factor of 2)
 * If the coefficient of the x-term is even


 * Solving Using the Quadratic Formula**
 * Always works, but is best used when the leading coefficient is 2 or greater **AND** you cannot factor


 * An additional fifth method that can be used to solve a quadratic equation is graphing. This will be explored in the next unit: Graphing Quadratic Functions.**


 * Let’s take a look at a few quadratic equations and determine the most efficient method for solving each using the criteria above.**

Equation 1: x 2 – 8x + 2 = 0
 * Solve by Factoring? Not efficient: although the leading coefficient is 1, there are no factors of 2 that would sum to -8
 * Solve Using Square Roots? Not efficient: there is an “x” term
 * Solve by Completing the Square? **Most efficient**: cannot be factored, the leading coefficient is 1 and the coefficient of the x-term is even
 * Solving Using the Quadratic Formula? Not efficient: The leading coefficient is not 2 or greater


 * Equation 2**: 2x 2 + 7x + 3 = 0
 * Solve by Factoring? **Most efficient:** the leading coefficient, 2, is a prime number, and you can factor using guess and check
 * Solve Using Square Roots? Not efficient: there is an “x” term
 * Solve by Completing the Square? Not efficient: the leading coefficient is not 1, and the coefficient of the x-term is not even
 * Solving Using the Quadratic Formula? Not efficient: Although the leading coefficient is 2 or greater, the trinomial can be factored using guess and check


 * Equation 3**: -5x 2 + 3x – 2 = 0
 * Solve by Factoring? Not efficient: although the leading coefficient is prime, this trinomial cannot be factored
 * Solve Using Square Roots? Not efficient: there is an “x” term
 * Solve by Completing the Square? Not efficient: the leading coefficient is not 1, and the coefficient of the x-term is not even
 * Solving Using the Quadratic Formula? **Most efficient**: The leading coefficient, 5, is 2 or greater, and the trinomial cannot be factored


 * Equation 4:** 3x 2 + 16 = 46
 * Solve by Factoring? Not efficient: although the leading coefficient is prime,
 * Solve Using Square Roots? **Most efficient**: there is an “x 2 ” term with no “x” term
 * Solve by Completing the Square? Not efficient: the leading coefficient is not 1, and there is no “x” term
 * Solving Using the Quadratic Formula? Not efficient: although the leading coefficient is 2 or greater, this can be solved by completing the square


 * Equation 5**: -5x 2 + 30x = 0
 * Solve by Factoring? **Most efficient**: you can factor out a GCF of -5x to eliminate the x 2 term
 * Solve Using Square Roots? Not efficient: there is an “x” term
 * Solve by Completing the Square? Not efficient: this can be factored and the leading coefficient is not 1
 * Solving Using the Quadratic Formula? Not efficient: although the leading coefficient is 2 or greater, this can be solved by factoring


 * Equation 6**: 2(x + 7) 2 – 14 = 32
 * Solve by Factoring? Not efficient: there is an (x + a) 2 in the equation
 * Solve Using Square Roots? **Most efficient**: there is an (x + a) 2 in the equation
 * Solve by Completing the Square? Not efficient: there is an (x + a) 2 in the equation
 * Solving Using the Quadratic Formula? Not efficient: there is an (x + a) 2 in the equation


 * Solving Quadratic Equations-Target B-Appropriate Quadratics-Quick Check**

Determine which method would be most appropriate for each of the following quadratic equations. Explain why.

1) 3x 2 + 10x – 5 = 0 2) 5(x – 4) 2 + 13 = 38 3) 6x 2 – 9x = 0 4) x 2 – 12x + 9 = 14

Solving Quadratic Equations-Target B-Appropriate Quadratics-Quick Check Solutions