Chapter 4: Q33. (page 163)

In Exercises 28-33 $x$ represents the length of a segment. When a value of $x$ doesn't make sense as a length, eliminate that value of $x$ .
$6 x^{2} = 5 x + 6$

Short Answer

Expert verified

The length of the segment is $x = 4$ .

Step by step solution

Step 1. Understand the concept used.

If $a x^{2} + b x + c = 0$ , with $a \neq 0$ , then $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.

Step 2. Substitute the values.

It is given that $6 x^{2} = 5 x + 6$ which can also be written as $6 x^{2} - 5 x - 6 = 0$ . Using quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , substitute 6 for a, $- 5$ for b and $- 6$ for c.

Put the values in the formula and get $x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \times (6) \times (- 6)}}{2 (6)}$ .

Step 3. Simplify for x.

Simplify the expression for x.

$x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \times (6) \times (- 6)}}{2 (6)} = \frac{5 \pm \sqrt{25 + 144}}{12} = \frac{5 \pm \sqrt{169}}{12} = \frac{5 \pm 13}{12}$

Therefore, either $x = \frac{5 + 13}{12} = \frac{18}{12}$ or $x = \frac{5 - 13}{12} = \frac{- 8}{12}$ that is., either $x = \frac{3}{2}$ or $x = - \frac{2}{3}$ . Here, the $x$ is having two values $\frac{3}{2}$ and $- \frac{2}{3}$ . Since, $x$ is a length of a segment that could not be negative or zero, therefore, the value of $x$ is $\frac{3}{2}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

In Exercises 28-33 $x$ represents the length of a segment. When a value of $x$ doesn't make sense as a length, eliminate that value of $x$ .
$6 x^{2} = 5 x + 6$

Short Answer

Step by step solution

Step 1. Understand the concept used.

Step 2. Substitute the values.

Step 3. Simplify for x.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Statistics

Pure Maths

Calculus

Decision Maths

Study anywhere. Anytime. Across all devices.

In Exercises 28-33xrepresents the length of a segment. When a value ofxdoesn't make sense as a length, eliminate that value ofx.6x2=5x+6

Short Answer

Step by step solution

Step 1. Understand the concept used.

Step 2. Substitute the values.

Step 3. Simplify for x.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Statistics

Pure Maths

Calculus

Decision Maths

Study anywhere. Anytime. Across all devices.

In Exercises 28-33 $x$ represents the length of a segment. When a value of $x$ doesn't make sense as a length, eliminate that value of $x$ .
$6 x^{2} = 5 x + 6$