/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Solve. \(6 x^{2}-23 x-55=0 \qu... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 55

Solve. \(6 x^{2}-23 x-55=0 \quad[3.2]\)

Short Answer

Expert verified
x = 5.5 and x = -\frac{5}{3}

Step by step solution

01

- Identify the coefficients

For the quadratic equation in the form of \(ax^2 + bx + c = 0\), identify the coefficients: \(a = 6\), \(b = -23\), and \(c = -55\).
02

- Apply the Quadratic Formula

The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substitute the values of \(a\), \(b\), and \(c\): \(a = 6\), \(b = -23\), and \(c = -55\).
03

- Calculate the Discriminant

Find the value under the square root (the discriminant): \(b^2 - 4ac = (-23)^2 - 4(6)(-55)\). Calculate it step-by-step: \((-23)^2 = 529\) and \(4 \times 6 \times -55 = -1320\). Put it together: \[529 - (-1320) = 529 + 1320 = 1849\].
04

- Solve for x using the Positive Root

Now calculate the two potential solutions. First, solve for the positive root: \[x = \frac{-(-23) + \sqrt{1849}}{2 \times 6} = \frac{23 + 43}{12} = \frac{66}{12} = 5.5\].
05

- Solve for x using the Negative Root

Now solve for the negative root: \[x = \frac{-(-23) - \sqrt{1849}}{2 \times 6} = \frac{23 - 43}{12} = \frac{-20}{12} = -\frac{5}{3}\].
06

- Provide the Final Solutions

The final solutions for \(x\) are: \(x = 5.5\) and \(x = -\frac{5}{3}\).

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Key Concepts

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

Quadratic Formula
Understanding the quadratic formula is essential in solving quadratic equations. A quadratic equation is typically written in the form ax2 + bx + c = 0, where a, b, and c are constants. The quadratic formula helps us find the roots of the equation, which are the values of x that make the equation true. The quadratic formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula can be used when factoring is difficult or impossible. It's vital to remember that the formula involves a square root, which means you will need to calculate the value under the square root (the discriminant) first.
Discriminant
The discriminant is a key part of the quadratic formula and is found under the square root: \[b^2 - 4ac\]Depending on its value, the discriminant tells us the nature of the roots:
  • If b2 - 4ac is positive, there are two distinct real roots.
  • If it is zero, there is exactly one real root (also called a repeated or double root).
  • If it is negative, there are no real roots, but two complex roots.
In our previous example, we calculated the discriminant to be 1849, which is positive. This means the quadratic equation 6x2 - 23x - 55 = 0 has two distinct real roots.Understanding the discriminant can save you time by giving a preview of the equation's solutions.
Roots of Quadratic Equation
Finding the roots of a quadratic equation is the final goal. Once you've used the quadratic formula and calculated the discriminant, you can solve for the roots by evaluating:\[x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\]and \[x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\]These two solutions correspond to the positive and negative roots. In the example given, we solved:
  • For the positive root: \[x = \frac{23 + 43}{12} = 5.5\],
  • For the negative root: \[x = \frac{23 - 43}{12} = -\frac{5}{3}\].
So, the roots or solutions to the equation are x = 5.5 and x = -\frac{5}{3}. By finding these roots, we effectively solve the quadratic equation.Understanding these roots gives you the answer to where the graph of the quadratic equation touches the x-axis.

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Most popular questions from this chapter

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Fill in the blank with the correct term. Some of the given choices will not be used. Others will be used more than once. $$\begin{array}{ll}x \text { -intercept } & \text { midpoint formula } \\ y \text { -intercept } & \text { horizontal lines } \\ \text { odd function } & \text { vertical lines } \\ \text { even function } & \text { point-slope equation } \\ \text { domain } & \text { slope-intercept equation } \\ \text { range } & \text { difference quotient } \\ \text { slope } & f(x)=f(-x) \\\ \text { distance formula } & f(-x)=-f(x)\end{array}$$ _____ are given by equations of the type \(x=a\).

Fill in the blank with the correct term. Some of the given choices will not be used. Others will be used more than once. $$\begin{array}{ll}x \text { -intercept } & \text { midpoint formula } \\ y \text { -intercept } & \text { horizontal lines } \\ \text { odd function } & \text { vertical lines } \\ \text { even function } & \text { point-slope equation } \\ \text { domain } & \text { slope-intercept equation } \\ \text { range } & \text { difference quotient } \\ \text { slope } & f(x)=f(-x) \\\ \text { distance formula } & f(-x)=-f(x)\end{array}$$ The _______ is \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).

List the critical values of the related function. Then solve the inequality. $$4 \geq \frac{4}{x}+x$$
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