Question
upstudy study bank question image url

12. Given that the roots of the equation \( x^{2}-21 x+4=0 \) are \( \alpha^{2} \) and \( \beta^{2} \), where both \( \alpha \) and \( \beta \) are positive, find (i) \( \alpha \beta \). (ii) \( \alpha+\beta \). (iii) an equation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \). 13. Let \( f(x)=x^{3}-5 x^{2}+2 x+8 \) be a polynomial.

Ask by Gordon Ross. in Zambia
Mar 13,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

**12.** Given the quadratic equation \( x^2 - 21x + 4 = 0 \) with roots \( \alpha^2 \) and \( \beta^2 \): **(i) \( \alpha \beta = 2 \)** **(ii) \( \alpha + \beta = 5 \)** **(iii) The equation with roots \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \) is \( 2x^2 - 21x + 2 = 0 \)** --- **13.** The polynomial \( f(x) = x^3 - 5x^2 + 2x + 8 \) factors as: \[ f(x) = (x + 1)(x - 2)(x - 4). \]

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Bonus Knowledge

To find \(\alpha \beta\), we can use Vieta's formulas. For the equation \(x^2 - 21x + 4 = 0\), the product of the roots \(\alpha^2 \beta^2\) is given by \(\frac{4}{1} = 4\). Thus, \(\alpha \beta = \sqrt{4} = 2\). Next, for \(\alpha + \beta\), we take the sum of the roots given by Vieta’s formulas again. The sum of the roots \(\alpha^2 + \beta^2\) is \(21\). We can relate this to \(\alpha + \beta\) using the identity \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha \beta\). So, substituting the values we get: \((\alpha + \beta)^2 = 21 + 2 \cdot 2 = 25\). Therefore, \(\alpha + \beta = \sqrt{25} = 5\). To find an equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\), we first note that these roots can also be expressed using simple transformations of Vieta’s formulas. The product of the roots is \(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1\) and the sum is \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{21}{2}\). Thus, the equation can be written as \(x^2 - Sx + P = 0\), or \(x^2 - \frac{21}{2}x + 1 = 0\). Multiplying through by 2 gives \(2x^2 - 21x + 2 = 0\). As for \(f(x)\), the polynomial \(f(x) = x^3 - 5 x^2 + 2 x + 8\) has real roots, and you can further explore its behavior using the Rational Root Theorem and synthetic division to factor or find numerically relevant information about its roots. How cool is it that cubic polynomials can have three real roots or a mix of real and complex?! Exploring graphs or using calculus can make this adventure even more thrilling!

Related Questions

Latest Algebra Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy