Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 194

For equation \(2 \mathrm{x}^{2}+16 \mathrm{x}+3 \mathrm{k}=0\) sum of the squares of roots is 10 then \(\mathrm{k}=\) (a) 12 (b) 15 (c) 18 (d) 21

Short Answer

Expert verified
The value of k is 18, which corresponds with answer choice (c).

Step by step solution

01

Find sum and product of roots

Following Vieta's formulas, for a quadratic equation of the form \(ax^2 + bx +c = 0\), the sum of its roots is given by \(-\frac{b}{a}\) and the product of its roots is given by \(\frac{c}{a}\). Here, our equation is \(2x^2 + 16x + 3k = 0\), where \(a = 2\), \(b = 16\), and \(c = 3k\). So, the sum of roots is: \(-\frac{16}{2} = -8\) And, the product of roots is: \(\frac{3k}{2}\) Let the roots be α and β. Thus, \(\alpha + \beta = -8\) and \(\alpha\beta = \frac{3k}{2}\).
02

Write sum of squares of roots in terms of α and β

We are given that the sum of squares of roots is 10. In terms of α and β, this can be expressed as: \(\alpha^2 + \beta^2 = 10\) We can rewrite the above expression in terms of the sum and product of roots: \((\alpha + \beta)^2 - 2\alpha\beta = 10\)
03

Substitute the values for sum and product of roots

Substitute the values of \(\alpha + \beta = -8\) and \(\alpha\beta = \frac{3k}{2}\) in the above equation: \((-8)^2 - 2\left(\frac{3k}{2}\right) = 10\) Now, we have an equation with k as the only unknown variable.
04

Solve the equation for k

Simplify the equation and solve for k: \(64 - 3k = 10\) Add 3k and subtract 10 from both sides: \(64 - 10 = 3k\) \(54 = 3k\) Divide by 3: \(k = 18\) The value of k is 18, which corresponds with answer choice (c).

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 solution set of equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]+16[x+(1 / x)]+26=0\) is (a) \([-1,\\{(-1) / 3\\},-3]\) (b) \([\overline{1,(1 / 3), 3}]\) (c) \([-1,(1 / 3), 3]\) (d) \([1,\\{(-1) / 3\\}, 3]\)

If \(\alpha, \beta\) are roots of equation \(a x^{2}+b x+c=0\) then value of \((\alpha a+b)^{-2}+(\beta a+b)^{-2}\) is (a) \(\left[\left(b^{2}-4 a c\right) /\left(a^{2} c^{2}\right)\right]\) (b) \(\left[\left(b^{2}-a c\right) /\left(a^{2} c^{2}\right)\right]\) (c) \(\left[\left(b^{2}-2 a c\right) /\left(a^{2} c^{2}\right)\right]\) (d) \(\left[\left(b^{2}+2 a c\right) /\left(a^{2} c^{2}\right)\right]\)

If \(\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}], \mathrm{x} \in \mathrm{R}-\\{0\\}\) where \([\mathrm{x}]=\) the greatest integer not greater than \(\mathrm{x}\), than number of solution \(\mathrm{f}(\mathrm{x})+\mathrm{f}(1 / \mathrm{x})=1 \ldots \ldots \ldots \ldots \ldots\) (a) 0 (b) 1 (c) 2 (d) infinite

For which value of \(b\), equation \(x^{2}+b x-1=0\) and \(\mathrm{x}^{2}+\mathrm{x}+\mathrm{b}=0\) have a common root. (a) \(-\sqrt{2}\) (b) \(-\mathrm{i} \sqrt{3}\) (c) \(\mathrm{i} \sqrt{5}\) (d) \(\sqrt{2}\)

If \(f(x)=2 x^{3}+m x^{2}-13 x+n\) and \((x-2),(x-3)\) are factor of \(\mathrm{f}(\mathrm{x})\) then \((\mathrm{m}, \mathrm{n})=\) (a) \((-5,-30)\) (b) \((-5,30)\) (c) \((5,30)\) (d) \((5,-30)\)
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Linguistic Terms

Read Explanation

Language and Social Groups

Read Explanation

Prosody

Read Explanation

International English

Read Explanation

Listening and Speaking

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