nicky442
Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

What are the zeros of 7x^2 -- 144 = --x^4 

please give an explanation and show the work. thank you!


Sagot :

Add x^4 to the left side and factor.
x^4 + 7x^2 - 144 = 0 then factor
(x^2 + 16)(x^2 - 9) = 0 set both factors = 0
x^2 + 16 = 0
x^2 = -16 square root each side
x = + and - 4i
x^2 - 9 = 0
x^2 = 9 square root each side
x = + and - 3
[tex]7x^2-144=-x^4\\\\x^4+7x^2-144=0\\\\(x^2)^2+7x^2-144=0\\\\substitute:t=x^2\geq0\\\\t^2+7t-144=0\\\\\Delta=7^2-4\cdot1\cdot(-144)=49+576=625;\ \sqrt\Delta=\sqrt{625}=25\\\\t_1=\frac{-7-25}{2\cdot1}=\frac{-32}{2}=-16 < 0;\ t_2=\frac{-7+25}{2\cdot1}=\frac{18}{2}=9\\\\x^2=9\iff x=\pm\sqrt9\to x=-3\ \vee\ x=3[/tex]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.