Make $y$ the subject in $ y^{2} = 3x^{2} - 2xy $

Question

How would I remove the $y$ in the right hand side so I can express $y$ in terms of x and visa versa? $$ Y^2 = 3X^2 - 2XY$$ I tried factoring but it’s not leading me anywhere. Any help is appreciated, thanks!

Answer 1

If you're stuck, you can always use the quadratic formula to solve for $y$. Treat $y$ as a variable and $x$ as a constant:

$$y^2 + 2xy - 3x^2 = 0$$ $$a = 1, b = 2x, c = -3x^2$$

and:

$$y = \frac{-b ± \sqrt{b^2-4ac}}{2a}$$

which leads to:

$$y = \frac{-2x ± \sqrt{4x^2-4(1)(-3x^2)}}{2}$$ $$= \frac{-2x ± \sqrt{16x^2}}{2}$$ $$ = -x ± 2x = -3x, x$$

Factorising also works. There aren't that many possibilities to consider, as the answer must be in the form $(y-a)(y+b) = 0$ or $(y+a)(y-b) = 0$, and $-ab = -3x^2$.

Answer 2

$$ Y^2+2XY-3X^2=0 $$ Now use the standard way of solving the quadratic equations. $$ Y=\frac{-2X\pm\sqrt{4X^2+9X^2}}{2}=-X\pm\frac{\sqrt{13}}{2}X = \left(-1\pm\frac{\sqrt{13}}{2}\right)X $$ To obtain $X$, you get this equation: $$ 3X^2-2XY-Y^2=0\\ X = \frac{2Y\pm\sqrt{4Y^2+9Y^2}}{2}=Y\pm\frac{\sqrt{13}}{2}Y = \left(1\pm\frac{\sqrt{13}}{2}\right)Y $$

Answer 3

From $$ y^2 = 3x^2 - 2xy, \tag{1} $$ we can write $$ y^2 + 2xy - 3x^2 = 0, $$ which can be treated as a quadratic equation in $y$ with solutions given by $$ y = \frac{-2x \pm \sqrt{ 4x^2 + 12x^2 } }{ 2 } = \frac{-2x \pm \sqrt{ 16 x^2} }{ 2 } = -x \pm 2 x = (-1 \pm 2)x. $$ Thus we have either $y = x$ or $y = -3x$.

And, similarly from (1) we can also write $$ 3x^2 - 2xy - y^2 = 0, $$ which is the same as $$ 3x^2 - 2yx - y^2 = 0, $$ which is a quadratic equation in $x$ with solutions given by $$ x = \frac{ 2y \pm \sqrt{ 4y^2 + 12y^2 } }{ 6 } = \frac{2 y \pm 4y}{2 y } = y \pm 2y = (1 \pm 2 ) y. $$ Thus we have either $x = 3y$ or $x = -y$.

Answer 4

The equation is $y^2= 3x^2- 2xy$. The first thing I would do is get all the "y"s on the left by adding 2xy to both sides: $y^2+ 2xy= 3x^2$. Now, instead of factoring or using the quadratic formula, I would complete the square. The coefficient of y is 2x. Half of that is x so we can "complete the square" on the left by adding $x^2$ to both sides

$y^2+ 2xy+ x^2= (y+ x)^2= 3x^2+ x^2= 4x^2$.

Now take the square root of both sides: $y+ x= \pm 2x$ so that either $y= 2x- x= x$ or $y= -2x- x= -3x$.

Check: if y= x then $y^2= x^2= 3x^2- 2x(x)= 3x^2- x^2= x^2$! If y= -3x then $y^2= 9x^2= 3x^2- 2x(-3x)= 3x^2+ 6x^2= 9x^2$!