# Real Roots of Exponential Equation (Involves Quadratic)

1. Sep 18, 2011

### seniorhs9

1. The problem statement, all variables and given/known data

Hi. I actually understand most of this question, but not the parts in red.

Question.

[PLAIN]http://img703.imageshack.us/img703/7237/2008testhphysf.jpg [Broken]

2. Relevant equations

The roots of a quadratic equation are real when $$b^2 - 4ac \geq 0$$

3. The attempt at a solution

Because we want real solutions, we have...

$$b^2 - 4ac \geq 0$$ so $$9 + 4k \geq 0 => \sqrt{9 + 4k} \geq 0$$ .

But $$y^2 - 3y + k = 0$$ has two solutions...

$$y_1 = \frac{1}{2}(3 - \sqrt{9 + 4k}$$

$$y_2 = \frac{1}{2}(3 + \sqrt{9 + 4k}$$

I get that $$9 + 4k \geq 0$$ means $$y_2 is true$$, but what about $$y_1$$?

The solution cares about "the larger root". Ie y_2. But why doesn't it care about y_1? I mean, both y_1, y_2 are solutions?

Thank you.

Last edited by a moderator: May 5, 2017
2. Sep 18, 2011

### dynamicsolo

3. Sep 18, 2011

### gb7nash

Context? Sometimes, when you're dealing with time problems, you'll get a positive and negative root, but since a negative time isn't valid, you'll just take the positive root.

However, if you're just given the equation:

$$y^2 -3y + k = 0$$

You are correct in saying that y1 and y2 are distinct real roots (assuming that 9+4k > 0).

4. Sep 18, 2011

### seniorhs9

Hi. I think I fixed my original post. The picture loads now.

gb7nash. Thanks for your answer. But this question and the test don't give any context for $$x$$ so I still don't see why the solution doesn't care about the negative root?

5. Sep 18, 2011

### gb7nash

Ok, I can see the image now.

Originally, we substitute y = 3x and we want to solve for y. Think about it for a second. Is there any way y can be negative? Is 3 raised to any power of x a negative number?

Knowing this now...

6. Sep 18, 2011

### seniorhs9

Actually, I understand why $$3^x > 0$$ for all x.

But my question is on how $$y_1 = \frac{1}{2}(3 - \sqrt{9 + 4k})$$ goes with the above.

We don't know $$3 - \sqrt{9 + 4k} > 0$$

We only know $$\sqrt{9 + 4k} \geq 0$$?

7. Sep 18, 2011

### gb7nash

You're exactly right. Depending on what value you choose for k, y1 may be positive or negative. For instance, for k = 10 you'll obtain a negative value for y1 (which is nonsense, since y cannot equal a negative number). For k = 0, you'll obtain a positive value for y1.

Since this is the case, we can't always rely on y1 to give a valid root, since there is the possibility of it being negative. However, y2 will always give a positive number (for k >= -9/4) and a valid root.

8. Sep 18, 2011

### seniorhs9

So are you saying the only final answer to this question is just y2, GIVEN $$\sqrt{9 + 4k} \geq 0$$?

I should toss out y1, since $$9 + 4k \geq 0$$ for this problem, because like we just talked about, y1 isn't always > 0 if $$9 + 4k \geq 0$$?

So just to make sure, y2 = 3x, GIVEN $$\sqrt{9 + 4k} \geq 0$$. AND it is WRONG to write y = 3x as final answer, if $$9 + 4k \geq 0$$.

Last edited: Sep 18, 2011
9. Sep 18, 2011

### gb7nash

No. A certain range of k makes y1 valid. What must k be to make y1 positive?

10. Sep 18, 2011

### seniorhs9

I edited my second last post because I actually didn't specify requirement $$9 + 4k \geq 0$$ for this problem. Now did I get everything right?

But to answer your question, y1 > 0 if $$3 - \sqrt{9 + 4k} > 0 <=> \sqrt{9 + 4k} < 3 <=> k < 0$$. But this isn't what we want for this question. Because if k < -9/4, original equation won't have real roots.

Last edited: Sep 18, 2011
11. Sep 18, 2011

### gb7nash

No, y1 > 0 if:

$$\sqrt{9+4k} < 3$$

Solving for k, what do you get?

Also, you want 9+4k > 0 in order to obtain a real root. Solve for k again.

Combining these, what must k be between to have two roots?

12. Sep 18, 2011

### seniorhs9

I actually meant to write $$3 - \sqrt{9 + 4k} > 0$$ so I've fixed this.

Because 3x = y > 0, the asked exponential question has AT LEAST two roots if

k < 0 and k > -9/4 <=> -9/4 < k < 0.

But I just want to make sure. The question asks "one or more real solutions", so isn't $$k \geq -9/4$$ enough?

This'd mean that it's enough to have y2 as the solution? y1 isn't necessary? Because "one" real solution's enough? Or is the answer wrong?

13. Sep 18, 2011

### gb7nash

Yes. If k >= -9/4, we can guarantee that we'll have one or two real roots (depending on if k is between -9/4 and 0, or greater than 0). I interpret one or more to mean that they want both cases covered, so k >= -9/4 gives you what you need.

14. Sep 18, 2011

### seniorhs9

So again, just to make sure, for only one real root, then

only y2 $$= \frac{1}{2}(3 + \sqrt{9 + 4k}$$ is a solution

so only y2 = 3x.

y1 $$= \frac{1}{2}(3 - \sqrt{9 + 4k}$$ is NOT a solution. I just want to make sure because I was confused about this.

15. Sep 19, 2011

### gb7nash

It depends on the value of k. As stated before, y1 and y2 are solutions for -9/4 <= k < 0, and only y2 is a solution for k >= 0. Since this is the case, a) is what you're looking for.