Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 175

If the ratio of the roots of the quadratic equation \(2 x^{2}+16 x+3 k=0\) is \(4: 5\) then \(k=\) (a) \([(2560) /(243)]\) (b) \(\overline{[(243) /(2560)]}\) (c) \([(-2560) /(243)]\) (d) \([(-243) /(2560)]\)

Short Answer

Expert verified
The correct answer is \(k= \frac{(-2560)}{243}\).

Step by step solution

01

Identify the sum and product of the roots

Given the quadratic equation \(2x^2 + 16x + 3k = 0\), let the roots be \(α\) and \(β\). According to the problem, the ratio of the roots is given by \(α:β = 4:5\). Thus, we can express the roots as \(α = 4m\) and \(β = 5m\), where \(m\) is a constant.
02

Apply Vieta's Formulas

Using Vieta's formulas, the sum and product of the roots can be expressed in terms of the coefficients of the quadratic equation as follows: Sum of the roots: \(α + β = \frac{-B}{A}\) Product of the roots: \(α × β = \frac{C}{A}\) Where A, B, and C are the coefficients of the quadratic equation \(2x^2 + 16x + 3k = 0\).
03

Calculate the sum and product of the roots

Using the expressions for the roots \(α = 4m\) and \(β = 5m\), we can calculate the sum and product of the roots: Sum of the roots: \(α + β = 4m + 5m = 9m\) Product of the roots: \(α × β = 4m × 5m = 20m^2\)
04

Set up and solve equations using Vieta's Formulas

Now, we will use Vieta's formulas to set up equations: Sum of the roots: \(\frac{-B}{A} = 9m\) ⟹ \(\frac{-16}{2} = 9m\) Product of the roots: \(\frac{C}{A} = 20m^2\) ⟹ \(\frac{3k}{2} = 20m^2\) Solving the first equation for \(m\): \(-8 = 9m\) \(m = -\frac{8}{9}\) Now, substitute the value of \(m\) in the second equation: \(\frac{3k}{2} = 20\left( -\frac{8}{9} \right)^2\)
05

Solve for k and find the correct answer

Now, we will solve the equation for the value of k: \(3k = 2 × 20\left( -\frac{8}{9} \right)^2\) \(k = \frac{2}{3} × 20\left( -\frac{8}{9} \right)^2\) \(k = \frac{(-2560)}{243}\) Comparing the calculated value of k with the given options, we can see that our answer matches option (c). Therefore, the correct answer is \(k= \boxed{\frac{(-2560)}{243}}\).

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The minimum value of \((\mathrm{x}+\mathrm{a})^{2}+(\mathrm{x}+\mathrm{b})^{2}+(\mathrm{x}+\mathrm{c})^{2}\) will be at \(\mathrm{x}\) equal to (a) \([(a+b+c) / 3]\) (b) \([\\{-(a+b+c)\\} / 3]\) (c) \(\sqrt{(a b c)}\) (d) \(a^{2}+b^{2}+c^{2}\)

If \(\alpha \& \beta\) are the roots of the equation \(x^{2}-x+1=0\) then \(\alpha^{2009}+\beta^{2009}=\) (a) \(-1\) (b) 1 (c) \(-2\) (d) 2

If the roots of the quadratic equation \((2 \mathrm{k}+3) \mathrm{x}^{2}+2(\mathrm{k}+3)\) \(x+(k+5)=0\\{k \in R, k \neq\\{(-3) / 2\\}\\}\) are equal, then \(\mathrm{K}=\) (a) 1,6 (b) \(-1,-6\) (c) \(-1,6\) (d) \(1,-6\)

If \(0

The value of \(\mathrm{k}\) for which the quadratic equation \(\mathrm{k} \mathrm{x}^{2}+1=\mathrm{k} \mathrm{x}+3 \mathrm{x}-11 \mathrm{x}^{2}\) has real and equal roots are (a) \(\\{-11,-3\\}\) (b) \(\\{5,7\\}\) (c) \(\\{5,-7\\}\) (d) \(\\{-5,-7\\}\)
See all solutions

Recommended explanations on English Textbooks

Phonology

Read Explanation

Listening and Speaking

Read Explanation

Phonetics

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Global English

Read Explanation

Research and Composition

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free