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
Solving Using Square Roots
Solving Using Completing the Square
Solving Using the Quadratic Formula
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.
In the examples below determine the most efficient method for solving each quadratic equation using the criteria above.
Example 1: \( x^2 – 8x + 2 = 0 \)
Example 2: \( 2x^2 + 7x + 3 = 0 \)
Example 3: \( -5x^2 + 3x – 2 = 0 \)
Example 4: \( 3x^2 + 16 = 46 \)
Example 5: \( -5x^2 + 30x = 0 \)
Example 6: \( 2(x + 7)^2 – 14 = 32 \)
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 “\( x2 \)” or quadratic term but no “\( x \)” or linear term
- If the equation has an expression that is of the form \( (x \pm h )^2 \)
Solving Using Completing the Square
- If you cannot factor the polynomial
- Always works, but easiest if the leading coefficient is 1 (or the leading coefficient is 1 after factoring out a factor of 2)and 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.
In the examples below determine the most efficient method for solving each quadratic equation using the criteria above.
Example 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
Example 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
Example 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
Example 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 “ \( x2 \) ” 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
Example 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
Example 6: \( 2(x + 7)^2 – 14 = 32 \)
- Solve by Factoring? Not efficient: there is an \( (x + h)^2 \) in the equation
- Solve Using Square Roots? Most efficient: there is an \( (x + h)^2 \) in the equation
- Solve by Completing the Square? Not efficient: there is an \( (x + h)^2 \) in the equation
- Solving Using the Quadratic Formula? Not efficient: there is an \( (x + h)^2 \) in the equation
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) \( x^2 – 12x + 9 = 14 \)
Quick Check Solutions
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) \( x^2 – 12x + 9 = 14 \)
Quick Check Solutions