x2+(y-3√2x)2=1 what is the value of this equation

Answers 1

Answer:

[tex]y=\sqrt{1-x^{2} } +3\sqrt{2x} \\y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]

Step-by-step explanation:

Given the equation is

[tex]x^{2} +(y-3\sqrt{2x} )^{2} =1[/tex]

Subtract [tex]x^{2}[/tex] from both sides of the equation

[tex](y-3\sqrt2{x} )^{2} +x^{2} -x^{2} =1-x^{2}[/tex]

Subtracting [tex]x^{2}[/tex] from itself leaves [tex]0[/tex]

[tex](y-3\sqrt{2x} )^{2} =1-x^{2}[/tex]

Take the square root of both sides of the equation

[tex]\sqrt{(y-3\sqrt{2x} )^{2} } =\sqrt{1-x^{2} }[/tex]

The square root operation always gives one positive number, and one negative number.

[tex](y-3\sqrt2{x} )=\sqrt{1-x^{2} } \\(y-3\sqrt{2x} )=-\sqrt{1-x^{2} }[/tex]

take[tex]-3\sqrt{2x}[/tex] on the right side then the value of [tex]y[/tex] is

[tex]y=\sqrt{1-x^{2} } +3\sqrt2{x}[/tex]

[tex]y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]

Then equation is solved and

[tex]y=\sqrt{1-x^{2} } +3\sqrt{2x} \\y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]