itsalizaeeeeyyyy
Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Show that the roots of the quadratic equation given by(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
are real and they can not be equal unless a=b=c

Sagot :

Medunno13

Expanding out the equation, we get

[tex]x^2 - ax - bx+ab+x^2 -bx-cx+bc+x^2 -ax-cx+ac=0 \\ \\ 3x^2 -x(2a+2b+2c)+(ab+bc+ac)=0[/tex]

Considering the discriminant,

[tex](2a+2b+2c)^2 - 4(3)(ab+bc+ac) \\ \\ =4a^2 + 4b^2 + 4c^2 + 8(ab+bc+ac)-12(ab+bc+ac) \\ \\ =4(a^2 + b^2 + c^2 - ab - bc - ac) \\ \\ =2((a-b)^2 + (b-c)^2 + (c-a)^2)[/tex]

This is always non-negative, meaning the roots are real.

The roots are equal if and only if the discriminant is 0, which is when a=b=c.

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.