Find Range of k for Two Positive Roots of Equation

[thereddevils]

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 ?

 

[annoymage]

(sorry wrong, i blatantly answer without reading the question ;P)

 

[D H]

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.

 

[thereddevils]

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 .

 

[D H]

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:

$$\aligned &10^{2x} - 10\,10^x + k = 0 \ \Rightarrow \\ &u^2 - 10u + k = 0 \qquad \text{with the substitution}\ u \equiv 10^x \endaligned$$

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

 

[thereddevils]

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:

$$\aligned &10^{2x} - 10\,10^x + k = 0 \ \Rightarrow \\ &u^2 - 10u + k = 0 \qquad \text{with the substitution}\ u \equiv 10^x \endaligned$$


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=10^x, 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 ?

 

[D H]

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=10^x.
• 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[/]?

 

[thereddevils]

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=10^x.
  
  
• 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

 

[D H]

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