Prove the Given Statement....2

[mathdad]

Suppose that the circle x^2 + 2Ax + y^2 + 2By = C has two intercepts, a and b, and two y-intercepts, c and d.
Prove that a•b - c•d = 0.

How is this started?

 

[MarkFL]

In the general quadratic:

\(\displaystyle ax^2+bx+c\)

Use the quadratic formula to obtain the product of the roots...what do you find?

 

[mathdad]

In the last example, the sum of the roots turned to be s = -b/a. Can the product p be -ba?

 

[MarkFL]

In the last example, the sum of the roots turned to be s = -b/a. Can the product p be -ba?

The product $P$ of the roots would be:

\(\displaystyle P=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=?\)

 

[mathdad]

I had a hunch that you were talking about multiplying the quadratic formula. I will be back later with the product.

 

[Evgeny.Makarov]

Mark is offering you a proof of Vieta's formulas (see example for application to quadratic polynomials).

 

[mathdad]

I know nothing about Vieta.

Here is the product P:

P = (4a^2b^2 + 4ac - b^2)/(4a^2)

What is the next step?

 

[Evgeny.Makarov]

I know nothing about Vieta.

That's why I provided a link. In my opinion, this information is useful, at least to verify the answers, i.e., formulas for sum and product of roots of a quadratic polynomial.

Here is the product P:

P = (4a^2b^2 + 4ac - b^2)/(4a^2)

No, using the formula $(x+y)(x-y)=x^2-y^2$ we get

\(\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{b^2-(b^2-4ac)}{4a^2}=\frac{4ac}{4a^2}=\frac{c}{a}\).

 

[mathdad]

What do I do with c/a?

 

[Evgeny.Makarov]

Going back to the original equation $x^2 + 2Ax + y^2 + 2By = C$ with $x$-intercepts $a$, $b$ and $y$-intercepts $c$, $d$, you can express $ab$ and $cd$ through $A$, $B$ and $C$ just like in the solved thread.

 

[mathdad]

Your typed letters are blocking your Latex work.

 

[Evgeny.Makarov]

Your typed letters are blocking your Latex work.

I am not sure what you mean (LaTeX formulas are also typed). If my post is displayed incorrectly, the best thing would be to post a screenshot to illustrate the problem.

 

[Greg]

I am not sure what you mean (LaTeX formulas are also typed). If my post is displayed incorrectly, the best thing would be to post a screenshot to illustrate the problem.

Inline latex throws RTCNTC's cell phone off (he does not use a desktop).

In regards to the problem,

Label the positive x-intercept a, the negative x-intercept b, the positive y-intercept c and the negative y-intercept d. These four points around the origin are the vertices of a cyclic orthodiagonal quadrilateral, so angle bdc is equivalent to angle bac and angle acd is equivalent to angle abd. Using the tangent ratio, b/d = c/a which implies ab = cd, ab - cd = 0; QED.

 

[mathdad]

Inline latex throws RTCNTC's cell phone off (he does not use a desktop).

In regards to the problem,

Label the positive x-intercept a, the negative x-intercept b, the positive y-intercept c and the negative y-intercept d. These four points around the origin are the vertices of a cyclic orthodiagonal quadrilateral, so angle bdc is equivalent to angle bac and angle acd is equivalent to angle abd. Using the tangent ratio, b/d = c/a which implies ab = cd, ab - cd = 0; QED.

Thanks for completing the problem. It's not that I do not use a desktop. I really do not have a computer or laptop. My cell phone is my desktop and laptop.

 

[Evgeny.Makarov]

In post #10 I wrote the following.

Going back to the original equation x^2 + 2Ax + y^2 + 2By = C with x-intercepts a, b and y-intercepts c, d, you can express ab and cd through A, B and C just like in the solved thread.

 

[MarkFL]

Suppose we wish to find the sum and product of the (distinct) roots of the general quadratic, and we know nothing of the quadratic formula. Let the two roots be $x_1$ and $x_2$. Since they are roots, we know:

\(\displaystyle ax_1^2+bx_1+c=0\)

\(\displaystyle ax_2^2+bx_2+c=0\)

Now, subtracting the former from the latter, we obtain:

\(\displaystyle a\left(x_2^2-x_1^2\right)+b\left(x_2-x_1\right)=0\)

Since the roots are distinct, then we may divide through by $x_2-x_1$ as it is not zero, and we obtain:

\(\displaystyle a\left(x_1+x_2\right)+b=0\implies x_1+x_2=-\frac{b}{a}\)

Now, if $c=0$, then we know the product of the roots is also 0. Then assuming neither root is 0, we may write:

\(\displaystyle ax_1+b=-\frac{c}{x_1}\)

\(\displaystyle ax_2+b=-\frac{c}{x_2}\)

Multiplying corresponding sides of these to equations together, we obtain:

\(\displaystyle a^2x_1x_2+ab\left(x_1+x_2\right)+b^2=\frac{c^2}{x_1x_2}\)

Using our formula for the sum of the roots, there results:

\(\displaystyle a^2x_1x_2-b^2+b^2=\frac{c^2}{x_1x_2}\)

\(\displaystyle a^2x_1x_2=\frac{c^2}{x_1x_2}\)

\(\displaystyle \left(x_1x_2\right)^2=\frac{c^2}{a^2}\)

Let's consider the case where the two roots have the same sign. Picture a parabola with both roots on the same side of the origin. We realize then that $a$ and $c$ must also have the same sign.

And likewise, when the two roots have opposite signs, then so too must $a$ and $c$ have opposite signs, and so we may write:

\(\displaystyle x_1x_2=\frac{c}{a}\)

 

[mathdad]

Thank you everyone. Great work, Mark!