Question

How do you solve the following by quadratic equation for `x` given `15x^2-16xy-15x^2=0`?

Answer 1

`x=(5/3)ycolor(white)("XX")orcolor(white)("XX")x=(-3/5)y`
(...but see below)

Explanation:

Since `15x^2-16xy-15y^2=0` would not normally be considered a quadratic equation, it is not obvious that this question is what what was really intended.

A quadratic equation is normally of the form:
`color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0`
We could treat the given equation as if it were in this form with
`color(white)("XXX")color(red)a = color(red)(15)`
`color(white)("XXX")color(blue)b=color(blue)(-16y)`
`color(white)("XXX")color(green)c=color(green)(-15y^2)`

The Quadratic Formula tells us
`color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)(b)^2-4color(red)acolor(green)c))/(2color(red)(a)`

This would give us
`color(white)("XXX")x=(16y+-sqrt(256y^2+900y^2))/(30)`

`color(white)("XXXX")=(16y+-2ysqrt(289))/(30)`

`color(white)("XXXX")=(8+-sqrt(289))/15y`

`color(white)("XXXX")=(8+-17)/15y`

`color(white)("XXXX")=25/15y=5/3ycolor(white)("XX")orcolor(white)("XX")-9/15y=-3/5y`

Answer 2

This is more of a case of providing information rather than solving the question. Unfortunately the image quality is not so good.
Using Maple I built the following:

Explanation:

In this image you can see the contour plot of that intersection.

Consider the cross section of the curved surface in the zy-plane. Observe that the vertex of that quadratic curve is considerably to the left of y=0.

Thus the termination point of the curves LHS has a significantly greater value for z than the RHS. Because of this, in the xy-plane view, the centre line of the saddle is sloping from the bottom left to the top right.

Consequently will influence the contour (implicitplot) of the line z=0 and produce the skewed affect.

In this image you can see the intersection of the plane `z=0`
Also displayed is the Maple solution for `15x^2-16xy-15y^2=0`
i.e. `z=0`

Maples solution reads:`[x=x" , " y=-5/3 x]" , "[x=5/3y" , "y=y]`

To demonstrate the solution output comparison to the contour plot have a look at:

This compares very well. Note that the saddle surface and the `z=0` plane would merge producing the strange looking gap near the centre of the contour plot.