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Mastering the Quadratic Formula for Solutions
Apr 8, 2025
Solving Quadratic Equations with the Quadratic Formula
Introduction
Learn to use the quadratic formula to solve quadratic equations.
Solutions should be given to two decimal places.
Quadratic Formula
Formula: ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )
Important: Memorize this formula for exams as it won't be provided.
Components of the Quadratic Formula
Standard form of a quadratic equation: ( ax^2 + bx + c = 0 )
( a ): Coefficient of ( x^2 )
( b ): Coefficient of ( x )
( c ): Constant term
Example 1: Solving a Quadratic Equation
Given Equation
( 2x^2 + 5x + 1 = 0 )
Steps to Solve
Identify ( a ), ( b ), and ( c ):
( a = 2 ), ( b = 5 ), ( c = 1 )
Substitute into the formula:
Substitute values into ( x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times 1}}{2 \times 2} )
Simplify inside the square root:
( x = \frac{-5 \pm \sqrt{25 - 8}}{4} )
( x = \frac{-5 \pm \sqrt{17}}{4} )
Calculate the solutions:
( x_1 = \frac{-5 + \sqrt{17}}{4} )
( x_2 = \frac{-5 - \sqrt{17}}{4} )
Approximate: ( x_1 = -0.22 ), ( x_2 = -2.28 )
Example 2: Solving a More Complex Quadratic Equation
Given Equation
( x^2 - 5x + 7 = 10 - 2x^2 )
Steps to Solve
Rearrange to standard form:
Move terms to get ( 3x^2 - 5x - 3 = 0 )
Identify ( a ), ( b ), and ( c ):
( a = 3 ), ( b = -5 ), ( c = -3 )
Substitute into the formula:
( x = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 3 \times (-3)}}{2 \times 3} )
Simplify inside the square root:
( x = \frac{5 \pm \sqrt{25 + 36}}{6} )
( x = \frac{5 \pm \sqrt{61}}{6} )
Calculate the solutions:
( x_1 = \frac{5 + \sqrt{61}}{6} )
( x_2 = \frac{5 - \sqrt{61}}{6} )
Approximate: ( x_1 = 2.14 ), ( x_2 = -0.468 )
Conclusion
Use the quadratic formula for non-factorable quadratics.
Check calculations with a calculator, especially for decimal or complex roots.
Practice substituting and simplifying to become efficient.
Recommendations
Practice with more equations to master the use of the quadratic formula.
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