Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
[tex]x+\frac{1}{x}=8\\\\\frac{x^2}{x}+\frac{1}{x}=8\\\\\frac{x^2+1}{x}=8\\\\x^2+1=8x\\\\x^2-8x+1=0[/tex]
[tex]a=1;\ b=-8;\ c=1\\\\\Delta=b^2-4ac\to\Delta=(-8)^2-4\cdot1\cdot1=64-4=60\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{60}=\sqrt{4\cdot15}=2\sqrt{15}\\\\x_1=\frac{8-2\sqrt{15}}{2\cdot1}=4-\sqrt{15};\ x_2=\frac{8+2\sqrt{15}}{2\cdot1}=4+\sqrt{15}\\\\Answer:4-\sqrt{15}\ or\ 4+\sqrt{15}.[/tex]
[tex]a=1;\ b=-8;\ c=1\\\\\Delta=b^2-4ac\to\Delta=(-8)^2-4\cdot1\cdot1=64-4=60\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{60}=\sqrt{4\cdot15}=2\sqrt{15}\\\\x_1=\frac{8-2\sqrt{15}}{2\cdot1}=4-\sqrt{15};\ x_2=\frac{8+2\sqrt{15}}{2\cdot1}=4+\sqrt{15}\\\\Answer:4-\sqrt{15}\ or\ 4+\sqrt{15}.[/tex]
You have said that x + 1/x = 8
Multiply each side by 'x' : x² + 1 = 8x
Subtract 8x from each side: x² - 8x + 1 = 0
Solving this quadratic equation by means of the quadratic formula,
(the procedure is left as an exercise for the student), the two
solutions are:
x = 4 + √15 = 7.873 (rounded)
and
x = 4 - √15 = 0.127 (rounded)
Multiply each side by 'x' : x² + 1 = 8x
Subtract 8x from each side: x² - 8x + 1 = 0
Solving this quadratic equation by means of the quadratic formula,
(the procedure is left as an exercise for the student), the two
solutions are:
x = 4 + √15 = 7.873 (rounded)
and
x = 4 - √15 = 0.127 (rounded)
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.