# Math Help - Three tough problems.

1. ## Three tough problems.

Sorry for such vague title.
Anyway, I am doing some math evaluation test when I ran across these bad boys.

For question #1, I have no idea where to begin.
For question #2, I forgot how to deal with absolute values.
For question #3:
$y= x+1$ So, $y^2 = x^2 + 2x +1$ Sub that into the other equation and we get $\frac{x^2}{4} + \frac{x^2 +2x+1}{9} = 1$
$9x^2 + 4x^2 + 8x + 4 = 36$
$13x^2 + 8x - 32 = 0$ but we do not get any nice roots, and all the options for the questions are rational.

If some one can show me how to solve these, I would appreciate it

2. ## Re: Three tough problems.

Hello, sakonpure61

$\text{The two curves }y \:=\:x+1\text{ and }\frac{x^2}{4} + \frac{y^2}{9} \:=\:1$

$\text{ intersect in exactly two points }(a,b)\text{ and }(c,d).$

$\text{Find the value of }a+c.$
$y\:=\: x+1 \quad\Rightarrow\quad y^2 \:=\: x^2 + 2x +1$

Sub that into the other equation and we get . $\frac{x^2}{4} + \frac{x^2 +2x+1}{9} \:=\: 1$

$9x^2 + 4x^2 + 8x + 4 \:=\: 36 \quad\Rightarrow\quad 13x^2 + 8x - 32 \:=\: 0$

But we do not get any nice roots, and all the options for the question are rational.

I bet your work is correct!

Quadratic Formula: . $x \:=\:\frac{\text{-}8 \pm\sqrt{1728}}{26} \:=\:\frac{\text{-}4\pm12\sqrt{3}}{13}$

The points are:
. . $(a,b) \:=\:\left(\frac{\text{-}4+12\sqrt{3}}{13},\:\frac{9+12\sqrt{3}}{13} \right)\;\text{ and }\;(c,d) \:=\:\left(\frac{\text{-}4-12\sqrt{3}}{13},\;\frac{9-12\sqrt{3}}{13}\right)$

$\text{Now add }a\text{ and }c\!: \;\frac{\text{-}4+12\sqrt{3}}{13} + \frac{\text{-}4 - 12\sqrt{3}}{13} \;=\;\cdots$

3. ## Re: Three tough problems.

Oh! I gave up too soon, thank you for the reply.

4. ## Re: Three tough problems.

Hey Sakon, hope that's a HOCKEY jersey in your avatar!!

5. ## Re: Three tough problems.

hehe, kind of an embarrassing situation, just look the other way ;D .

6. ## Re: Three tough problems.

The answer to 10 is B, and here I will teach you how to cheat lol.
If |b| > a > 0, that requires b>a or b<-a. Hence, it is impossible for b to stand between -a and a.
That's why B is incorrect. 6-x^2 cannot stay between -3 and 3 if |6-x^2| >=3.

7. ## Re: Three tough problems.

For the first question, I am gonna find out the equation for line BC first, and you shall see why.
slope: (0-1)/(1- -3) = -1/4, the line should be y=(-1/4)x+b. Plugging C(1, 0) in, you should get b= 1/4 so the line BC is y=(-1/4)x+(1/4)
Here's the thing: since HA and BC are perpendicular, THE PRODUCT OF THEIR SLOPES will be -1.
Hence, you know that the slope for HA is 4.
Then, y=4x+b for HA. Plugging A(-1, 2) in should get b= 6. Then line HA is y=4x+6.
Since H is the intersection point of BC and HA, that means y=4x+6=(-1/4)x+(1/4).
To solve that, I am gonna multiply both sides by 4: 16x+24 = -x+1 => 17x = -23, x=-23/17, y=4(-23/17) + 6 = 10/17.
A?