Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
[tex]15x^2-1=2x\\\\15x^2-2x-1=0\\\\a=15;\ b=-2;\ c=-1\\\\\Delta=b^2-4ac\to\Delta=(-2)^2-4\cdot15\cdot(-1)=4+60=64\\\\x_1=\frac{-b-\sqrt\Delta}{2}a;\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{64}=8\\\\x_1=\frac{2-8}{2\cdot15}=\frac{-6}{30}=-\frac{1}{5};\ x_2=\frac{2+8}{2\cdot15}=\frac{10}{30}=\frac{1}{3}[/tex]
[tex]15x^2 - 1 = 2x \\\\15x^2-2x - 1 = 0\\ \\a=15, \ \ b=-2 , \ \ c=-1 \\ \\\Delta =b^2-4ac = (-2)^2 -4\cdot 15 \cdot (-1) = 4+60=64 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-(-2)-\sqrt{64}}{2\cdot 15 }=\frac{ 2-8}{30}=\frac{-6}{30}=- \frac{1}{5}\\ \\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-(-2)+\sqrt{64}}{2\cdot 15 }=\frac{ 2+8}{30}=\frac{10}{30}= \frac{1}{3}
[/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.