Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Given
[tex]25x^{-4} - 99x^{-2} - 4 = 0[/tex]
consider substituting [tex]y=x^{-2}[/tex] to get a proper quadratic equation,
[tex]25y^2 - 99y - 4 = 0[/tex]
Solve for [tex]y[/tex] ; we can factorize to get
[tex](25y + 1) (y - 4) = 0[/tex]
[tex]25y+1 = 0 \text{ or } y - 4 = 0[/tex]
[tex]y = -\dfrac1{25} \text{ or }y = 4[/tex]
Solve for [tex]x[/tex] :
[tex]x^{-2} = -\dfrac1{25} \text{ or }x^{-2} = 4[/tex]
The first equation has no real solution, since [tex]x^{-2} = \frac1{x^2} > 0[/tex] for all non-zero [tex]x[/tex]. Proceeding with the second equation, we get
[tex]x^{-2} = 4 \implies x^2 = \dfrac14 \implies x = \pm\sqrt{\dfrac14} = \boxed{\pm \dfrac12}[/tex]
If we want to find all complex solutions, we take [tex]i=\sqrt{-1}[/tex] so that the first equation above would have led us to
[tex]x^{-2} = -\dfrac1{25} \implies x^2 = -25 \implies x = \pm\sqrt{-25} = \pm5i[/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
