How do you solve the following by quadratic equation for x given 15x2−16xy−15x2=0?

2 Answers
Alan P.
Oct 12, 2016

x=(53)yXXorXXx=(−35)y
(...but see below)

Explanation:

Since 15x2−16xy−15y2=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:
XXXax2+bx+c=0
We could treat the given equation as if it were in this form with
XXXa=15
XXXb=−16y
XXXc=−15y2

The Quadratic Formula tells us
XXXx=−b±√b2−4ac2a

This would give us
XXXx=16y±√256y2+900y230

XXXX=16y±2y√28930

XXXX=8±√28915y

XXXX=8±1715y

XXXX=2515y=53yXXorXX−915y=−35y

Tony B
Oct 13, 2016

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.
TonyB

In this image you can see the intersection of the plane z=0
Also displayed is the Maple solution for 15x2−16xy−15y2=0
i.e. z=0
Tony B

Maples solution reads:[x=x , y=−53x] , [x=53y , y=y]

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

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.

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