Problem 20 Exercises $1-28:$ Solve the qu... [FREE SOLUTION]

Chapter 3: Problem 20

Exercises $1-28:$ Solve the quadratic equation. Check your answers for Exercises $1-12$. $$ \frac{3}{4} x^{2}+\frac{1}{2} x-\frac{1}{2}=0 $$

Short Answer

Expert verified

The solutions are $ x = \frac{-1 + \sqrt{13}}{3} $ and $ x = \frac{-1 - \sqrt{13}}{3} $.

Step by step solution

Identify the coefficients

First, identify the coefficients from the given quadratic equation, which is \[ \frac{3}{4}x^{2} + \frac{1}{2}x - \frac{1}{2} = 0 \]The quadratic equation format is $ ax^2 + bx + c = 0 $. Thus:\[ a = \frac{3}{4}, \quad b = \frac{1}{2}, \quad c = -\frac{1}{2} \]

Calculate the discriminant

Calculate the discriminant using the formula:\[ D = b^2 - 4ac \]Substitute the values found:\[ D = \left(\frac{1}{2}\right)^2 - 4\left(\frac{3}{4}\right)\left(-\frac{1}{2}\right) \]Simplify the expression:\[ D = \frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{13}{4} \]

Apply the quadratic formula

Use the quadratic formula to find the roots:\[ x = \frac{-b \pm \sqrt{D}}{2a} \]Substitute the known values of $ b $, $ D $, and $ a $:\[ x = \frac{-\frac{1}{2} \pm \sqrt{\frac{13}{4}}}{2 \times \frac{3}{4}} \]Simplify:\[ x = \frac{-\frac{1}{2} \pm \frac{\sqrt{13}}{2}}{\frac{3}{2}} \]Multiply numerator and denominator by 2:\[ x = \frac{-1 \pm \sqrt{13}}{3} \]

State the solutions

From the quadratic formula, the solutions are:\[ x_1 = \frac{-1 + \sqrt{13}}{3} \]\[ x_2 = \frac{-1 - \sqrt{13}}{3} \]

Verify the solutions

Substitute the solutions back into the original equation to verify.For $ x_1 $:Replace $ x $ with $ \frac{-1 + \sqrt{13}}{3} $ in the original equation and simplify to ensure it satisfies the equation.For $ x_2 $:Replace $ x $ with $ \frac{-1 - \sqrt{13}}{3} $ in the original equation and simplify to ensure it satisfies the equation.Both should equal zero, confirming the solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discriminant

In the realm of quadratic equations, the discriminant plays a crucial role. It helps determine the nature of the roots without actually solving the equation.
The discriminant (denoted as $ D $) is calculated using the formula:

$ D = b^2 - 4ac $

Here, $ a $, $ b $, and $ c $ are the polynomial coefficients from the quadratic equation of the form $ ax^2 + bx + c = 0 $.
The discriminant gives insights into the type of roots:

If $ D > 0 $, the equation has two distinct real roots.
If $ D = 0 $, the equation has one real root, often called a repeated or double root.
If $ D < 0 $, the equation has two complex roots.

In the given exercise, substituting the values $ b = 1/2 $, $ a = 3/4 $, and $ c = -1/2 $ into the formula gives $ D = 13/4 $. Since $ D > 0 $, the equation has two real roots.

Quadratic Formula

The quadratic formula is a straightforward and reliable method to find the roots of any quadratic equation.
This powerful formula is:

$ x = \frac{-b \pm \sqrt{D}}{2a} $

Here, $ D $ stands for the discriminant ($ b^2 - 4ac $), and $ a $, $ b $, and $ c $ are the coefficients of the quadratic equation.
When solving a quadratic equation:1. First calculate the discriminant $ D $.2. Substitute $ b $, $ a $, and the calculated $ \sqrt{D} $ into the quadratic formula.
For the equation $ \frac{3}{4}x^2 + \frac{1}{2}x - \frac{1}{2} = 0 $,

$ b = 1/2 $
$ a = 3/4 $
$ D = 13/4 $
Plugging these into the formula gives the two solutions:
$ x_1 = \frac{-1 + \sqrt{13}}{3} $
$ x_2 = \frac{-1 - \sqrt{13}}{3} $

Polynomial Coefficients

Polynomial coefficients are the numbers $ a $, $ b $, and $ c $ in a quadratic equation $ ax^2 + bx + c = 0 $.
These coefficients play a significant part in controlling the shape and position of a parabola, which is the graph of a quadratic equation.

$ a $ is the quadratic coefficient and affects the direction and width of the parabola:

If $ a > 0 $, the parabola opens upwards.
If $ a < 0 $, the parabola opens downwards.
Larger $ |a| $ values result in a narrower parabola.

$ b $ is the linear coefficient, influencing the tilt or the axis intercepts of the parabola.
$ c $ is the constant term, representing the y-intercept where the graph intersects the y-axis.

In the original exercise $ \frac{3}{4}x^2 + \frac{1}{2}x - \frac{1}{2} = 0 $, the coefficients are:

$ a = \frac{3}{4} $
$ b = \frac{1}{2} $
$ c = -\frac{1}{2} $

Understanding these coefficients provides a deeper insight into solving quadratic equations and predicting the parabolic graph's properties.

91影视

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ \frac{3}{4} x^{2}+\frac{1}{2} x-\frac{1}{2}=0 $$

Short Answer

Step by step solution

Identify the coefficients

Calculate the discriminant

Apply the quadratic formula

State the solutions

Verify the solutions

Key Concepts

Discriminant

Quadratic Formula

Polynomial Coefficients

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