Question

x²+(y-3√2x)²=1 slove this

Answer 1

Answer:

I really hope this helps you understand the concept of graphing!

Step-by-step explanation:

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Answer 2

We need to recall the following formulas.

• Quadratic formula: The roots of the quadratic equation $ax^2+bx+c=0$ are $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$
• $(a-b)^2=a^2+b^2-2ab$

Given:

$x^{2} +(y-3\sqrt{2}x)^ 2=1$

Simplify the equation.

$x^{2} +y^{2}+18x^2-6\sqrt{2}yx=1$

$19x^{2} -(6\sqrt{2}y)x+(y^{2}-1)=0$

Using the quadratic formula, we get

$x_{1,2}=\frac{-(-6\sqrt{2}y) \pm\sqrt{(-6\sqrt{2}y)^2-4(19)(y^2-1)} }{2(19)}$

$x_{1,2}=\frac{6\sqrt{2}y \pm\sqrt{72y^2-76y^2+76} }{38}$

$x_{1,2}=\frac{6\sqrt{2}y \pm\sqrt{-4y^2+76} }{38}$

$x_{1,2}=\frac{6\sqrt{2}y \pm2\sqrt{-y^2+19} }{38}$

$x_{1,2}=\frac{3\sqrt{2}y \pm\sqrt{-y^2+19} }{19}$

Hence, the values of $x$ are as follows

$x=\frac{3\sqrt{2}y +\sqrt{-y^2+19} }{19}$   or  $x=\frac{3\sqrt{2}y -\sqrt{-y^2+19} }{19}$