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

Oct 12, 2016

$x = \left(\frac{5}{3}\right) y \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = \left(- \frac{3}{5}\right) y$
(...but see below)

#### Explanation:

Since $15 {x}^{2} - 16 x y - 15 {y}^{2} = 0$ 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:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
We could treat the given equation as if it were in this form with
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} = \textcolor{red}{15}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{b} = \textcolor{b l u e}{- 16 y}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{c} = \textcolor{g r e e n}{- 15 {y}^{2}}$

color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)(b)^2-4color(red)acolor(green)c))/(2color(red)(a)

This would give us
$\textcolor{w h i t e}{\text{XXX}} x = \frac{16 y \pm \sqrt{256 {y}^{2} + 900 {y}^{2}}}{30}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{16 y \pm 2 y \sqrt{289}}{30}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{8 \pm \sqrt{289}}{15} y$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{8 \pm 17}{15} y$

$\textcolor{w h i t e}{\text{XXXX")=25/15y=5/3ycolor(white)("XX")orcolor(white)("XX}} - \frac{9}{15} y = - \frac{3}{5} y$

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. In this image you can see the intersection of the plane $z = 0$
Also displayed is the Maple solution for $15 {x}^{2} - 16 x y - 15 {y}^{2} = 0$
i.e. $z = 0$ Maples solution reads:$\left[x = x \text{ , " y=-5/3 x]" , "[x=5/3y" , } y = y\right]$

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.