Real Roots of Exponential Equation (Involves Quadratic)

[seniorhs9]

Homework Statement

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

Question.

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

If above doesn't load, please go to http://img703.imageshack.us/img703/7237/2008testhphysf.jpg

Homework Equations

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

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.

 

[dynamicsolo]

Your attachment isn't coming through.

 

[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).

 

[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?

 

[gb7nash]

Ok, I can see the image now.

Originally, we substitute y = 3^x 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...

 

[seniorhs9]

Hi gb7nash. Thanks for your answer.

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?

 

[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.

 

[seniorhs9]

Hi gb7nash. Thanks for your answer.

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 = 3^x, GIVEN \sqrt{9 + 4k} \geq 0. AND it is WRONG to write y = 3^x as final answer, if 9 + 4k \geq 0.

 

[gb7nash]

So are you saying the only final answer to this question is just y₂?

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

 

[seniorhs9]

Hi gb7nash. Thanks for your answer.

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} &gt; 0 &lt;=&gt; \sqrt{9 + 4k} &lt; 3 &lt;=&gt; k &lt; 0. But this isn't what we want for this question. Because if k < -9/4, original equation won't have real roots.

 

[gb7nash]

No, y₁ > 0 if:

\sqrt{9+4k} &lt; 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?

 

[seniorhs9]

Hi gb7nash. Thanks for your answer.

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

Because 3^x = 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?

 

[gb7nash]

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

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.

 

[seniorhs9]

Hi gb7nash. Thank you so much for your answers.

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 = 3^x.

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

 

[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.