Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
[tex]y^2-6y+6=0\\
y^2-6y+9-3=0\\
(y-3)^2=3\\
|y-3|=\sqrt3\\
y-3=\sqrt3 \vee y-3=-\sqrt3\\
y=3+\sqrt3 \vee y=3-\sqrt3[/tex]
[tex]y^2-6y+6=0\\ \\a=1 , \ b=-6, \ c=6 \\ \\\Delta =b^2-4ac = (-6)^2 -4\cdot1\cdot6 = 36-24=12\\ \\\sqrt{\Delta }= \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3} \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{6-2\sqrt{3}}{2 }=\frac{2( 3- \sqrt{3})}{2}= 3- \sqrt{3}[/tex]
[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{6+2\sqrt{3}}{2 }=\frac{2( 3+ \sqrt{3})}{2}= 3+ \sqrt{3}[/tex]
[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{6+2\sqrt{3}}{2 }=\frac{2( 3+ \sqrt{3})}{2}= 3+ \sqrt{3}[/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.