Solve by using the quadratic formula. Prove that your answer is correct by checking the...

Question:

Solve by using the quadratic formula. Prove that your answer is correct by checking the solutions.

{eq}2x^2 + 3x +4= 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 roots (solutions) to this type of equation are formed by the values of the variable x, 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} + 3x + 4 = 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 = 3,\,{\text{ and }}\,c = 4.{\text{ Thus:}} \cr & \,\,\,\,\,x = \frac{{ - 3 \pm \sqrt {{3^2} - 4 \cdot 2 \cdot 4} }}{{2 \cdot 2}} \cr & {\text{Simplifying:}} \cr & \,\,\,\,\,x = \frac{{ - 3 \pm \sqrt {23} i}}{4} \cr & {\text{Thus:}} \cr & \,\,\,\,\, \Rightarrow x = - \frac{3}{4} + \frac{{\sqrt {23} }}{4}i,\,\,\,\,\,\,\,x = - \frac{3}{4} - \frac{{\sqrt {23} }}{4}i \cr & {\text{Therefore}}{\text{, the solutions for this equation are: }} \cr & \,\,\,\,\,\boxed{x = - \frac{3}{4} + \frac{{\sqrt {23} }}{4}i,\,\,\,\,x = - \frac{3}{4} - \frac{{\sqrt {23} }}{4}i} \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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