Solve and Find the exact solution(s). Simplify as much as possible. Show work. 2x^2 + 4x -1 = 0

Question:

Solve and Find the exact solution(s). Simplify as much as possible. Show work.

{eq}2x^2 + 4x-1 = 0 {/eq}

Quadratic Equation:

A quadratic equation is easily recognizable, since it is a second-degree equation of the polynomial type, and it has the following simplified form: {eq}a{x^2} + bx + c = 0 {/eq}. The solutions to this type of equation are formed by the values of the variable {eq}x {/eq}, for which the quadratic expression is equal to 0 and is written as: {eq}\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} {/eq}.

Answer and Explanation:

{eq}\eqalign{ & {\text{We have the quadratic function }}\,2{x^2} + 4x - 1 = 0.{\text{ }} \cr & {\text{So}}{\text{, for a quadratic equation of the form }}a{x^2} + bx + c = 0\,{\text{ }} \cr & {\text{the solutions (roots) are given by the quadratic formula:}} \cr & \,\,\,\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \cr & {\text{In this particular case }}\,a = 2,\,\,\,b = 4,\,{\text{ and }}\,c = - 1.{\text{ Thus:}} \cr & \,\,\,\,\,x = \frac{{ - 4 \pm \sqrt {{4^2} - 4 \cdot 2 \cdot \left( { - 1} \right)} }}{{2 \cdot 2}} \cr & {\text{Simplifying:}} \cr & \,\,\,\,\,x = \frac{{ - 4 \pm \sqrt {24} }}{4} \cr & \,\,\,\,\,x = \frac{{ - 4 \pm 2\sqrt 6 }}{4} \cr & \,\,\,\,\,x = \frac{{ - 2 \pm \sqrt 6 }}{2} \cr & {\text{Thus:}} \cr & \,\,\,\,\, \Rightarrow x = \frac{{ - 2 + \sqrt 6 }}{2},\,\,\,\,\,\,\,x = \frac{{ - 2 - \sqrt 6 }}{2} \cr & {\text{Therefore}}{\text{, the solutions for this equation are: }}\,\boxed{x = \frac{{ - 2 + \sqrt 6 }}{2},\,\,\,\,x = \frac{{ - 2 - \sqrt 6 }}{2}} \cr} {/eq}


Learn more about this topic:

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How to Use the Quadratic Formula to Solve a Quadratic Equation

from Math 101: College Algebra

Chapter 4 / Lesson 10
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