MHB Prove the Given Statement....1

[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) = A/B.

How is this started?

 

[MarkFL]

Suppose we are given the general quadratic:

$$ax^2+bx+c=0$$

Now the quadratic formula tells us the roots are:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

And so the sum of the roots $s$ is:

$$s=-\frac{b}{a}$$

In the given problem, we are told that the $x$-intercepts of the circle are $a$ and $b$. This means at these points $y=0$ and so we are left with:

$$x^2+2Ax-C=0$$

We know the roots are $a$ and $b$, and so the sum of the roots is:

$$a+b=-\frac{2A}{1}=-2A$$

What about the $y$-intercepts?

 

[mathdad]

By sum of the roots you mean ADDING a positive and negative quadratic formula, right?

 

[mathdad]

For the y-intercept, x must be 0.

x^2 + 2Ax + y^2 + 2By = C

(0)^2 + 2A(0) + y^2 + 2By = C

y^2 + 2By - C = 0

Let s = sum of roots

s = -b/a

c + d = -2B/1 = -2BProve that (a + b)/(c + d) = A/B.

a + b = -2A

c + d = -2B

-2A/-2B = A/B

Done.