Find Range of k for Two Positive Roots of Equation

AI Thread Summary
The discussion revolves around finding the range of k for the equation 100^x - 10^(x+1) + k = 0 to have two distinct positive roots. The initial conclusion was that k must be less than 25 for two distinct roots, but further analysis revealed that k must also be greater than 0 to ensure both roots are positive. The substitution of u = 10^x clarified that u must be greater than 0, leading to the conclusion that the correct range for k is 0 < k < 25. Participants emphasized the importance of distinguishing between real and positive roots in their calculations. The final consensus is that the range of k ensuring two distinct positive roots is indeed 0 < k < 25.
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Homework Statement



Find the range of the number k so that the equation 100^x-10^(x+1)+k=0 has two distinct positive roots

Homework Equations





The Attempt at a Solution



I know if it says for two distinct roots only , k<25

but now its two distinct POSITIVE roots , so how ?
 
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(sorry wrong, i blatantly answer without reading the question ;P)
 
Last edited:
One way to solve this problem is to find the generic solutions and then find the range of k that makes both solutions positive.

How did you arrive at k<25? You must have done some work on this problem to arrive at that value. Show it.
 
D H said:
One way to solve this problem is to find the generic solutions and then find the range of k that makes both solutions positive.

How did you arrive at k<25? You must have done some work on this problem to arrive at that value. Show it.

ok .

10^(2x)-10^x . 10 + k=0

Let 10^x be b

b^2-10b+k=0

Then since it says two +ve distinct roots ,

b^2-4ac>0 , which is how i found k<25

but the answer given is 0<k<25

I don see where is the 0 coming , i know its sth to do with the word positive .
 
thereddevils said:
ok .

10^(2x)-10^x . 10 + k=0

Let 10^x be b

b^2-10b+k=0
Ok so far, but your choice of b here is going to get you in trouble. Something like u would have been a much better choice:

\aligned<br /> &amp;10^{2x} - 10\,10^x + k = 0 \ \Rightarrow \\<br /> &amp;u^2 - 10u + k = 0 \qquad \text{with the substitution}\ u \equiv 10^x<br /> \endaligned

b^2-4ac>0 , which is how i found k<25
This is where the choice of b will get you in trouble. This b is not the same as your variable b.

Regarding the problem itself: Are you supposed to find the range of k that yields two distinct real solutions for x, or the range that yields two positive solutions?Hint: with u=10x, what values for u correspond to a real value for x? To a positive value of x?
 
D H said:
Ok so far, but your choice of b here is going to get you in trouble. Something like u would have been a much better choice:

\aligned<br /> &amp;10^{2x} - 10\,10^x + k = 0 \ \Rightarrow \\<br /> &amp;u^2 - 10u + k = 0 \qquad \text{with the substitution}\ u \equiv 10^x<br /> \endaligned


This is where the choice of b will get you in trouble. This b is not the same as your variable b.

Regarding the problem itself: Are you supposed to find the range of k that yields two distinct real solutions for x, or the range that yields two positive solutions?


Hint: with u=10x, what values for u correspond to a real value for x? To a positive value of x?


ok ,

u^2 - 10u + k = 0

then b^2-4ac>0

(-10)^2-4k>0

100-4k>0

k<25 , still i get the same thing .

And this range of values of k is supposed two distinct , positive and real solution . Is it possible ?
 
You are too hung up on the range.

Some questions,
  • What are the solutions in terms of u?
    This is a simple quadratic equation.
  • What is x in terms of u?
    Forget about the quadratic equation in answering this question. All that matters is that u=10x.
  • How do these solutions for u translate to solutions for x?
    Use the above.
  • Does the answer to the above question always make sense?
    Just because there are two real solutions for u does not necessarily mean these translate to two real solutions for x.
  • What is the range of u that yields positive values for x?
    Once again forget about the quadratic equation in answering this question.
  • What does that translate to in terms of k[/]?
 
D H said:
You are too hung up on the range.

Some questions,
  • What are the solutions in terms of u?
    This is a simple quadratic equation.

  • What is x in terms of u?
    Forget about the quadratic equation in answering this question. All that matters is that u=10x.

  • How do these solutions for u translate to solutions for x?
    Use the above.

  • Does the answer to the above question always make sense?
    Just because there are two real solutions for u does not necessarily mean these translate to two real solutions for x.

  • What is the range of u that yields positive values for x?
    Once again forget about the quadratic equation in answering this question.

  • What does that translate to in terms of k[/]?


thanks ! Or maybe i can also do it this way ,

since i make the substitution u=10^x , u>0

u^2-10u+k=0

so k>0 -- 1

Then since it has two distinct roots , b^2-4ac>0 which implies k<25 ---2

so combining 1 and 2

0<k<25
 
No what about two positive roots, or was a misstatement in the original post?
 
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