Exponential Function

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danielvnl

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What real values of m the equation 4x - (m-2)*2x + 2m +1 = 0 admit at least one real root ?

I'd like to know what properties I have to know to solve this problem and what roles of each ones take in this question.

Let's go slowly ...

I'm trying 2x = y whe have y2 - (m-2)y + (2m+1) = 0

here foward I don't know what to do, I'm trying to understand what my book is saying, but I'm having some difficulties. Let's see where...

" The equation admits at least one real root if y = 2x > 0 "

Why ??
Is y = 22 > 0 a property ??
What happens if y = 2x < 0 ??
 
danielvnl said:
What real values of m the equation 4x - (m-2)*2x + 2m +1 = 0 admit at least one real root ?

I'd like to know what properties I have to know to solve this problem and what roles of each ones take in this question.

Let's go slowly ...

I'm trying 2x = y whe have y2 - (m-2)y + (2m+1) = 0

here foward I don't know what to do, I'm trying to understand what my book is saying, but I'm having some difficulties. Let's see where...

" The equation admits at least one real root if y = 2x > 0 "

Why ??
Is y = 22 > 0 a property ??
What happens if y = 2x < 0 ??
Click to expand...

Don't you see that \(\displaystyle 2^x\not<0~?\)
 
pka

Sorry, I haven't seen that, so obvious. Whith this we have roots > 0, so we have two possibilities

1 - two positives roots, with

math.JPG

which intersection is m >= 12

2 - only one positive root, with
math3.JPG

m < -1/2


aswer

math2.JPG

Thx very much ...
 
Is sure the answer of this question is

m < -1/2 or m >= 12

but I thought there where another way to solve this.
 
danielvnl said:
Is sure the answer of this question is

m < -1/2 or m >= 12

but I thought there where another way to solve this.
Click to expand...

danielvnl, as you had it in the first post, it can be rewritten as \(\displaystyle y^2 - (m - 2)y + (2m + 1) = 0, \ \ using \ \ 2^x = y. \ \ \) If the quadratic formula is used, then \(\displaystyle \ \ y \ = \ \dfrac{m - 2 \pm\sqrt{m^2 - 12m \ }}{2}. \ \ \ \) From that, there are boundaries by which the m-values can be found.
 
Hello, danielvnl!

You were off to a good start . . .

\(\displaystyle \text{For what real values of }m\text{ does the equation }\:4^x - (m-2)2^x + 2m +1 \:=\:0\)
. . \(\displaystyle \text{ have at least one real root?}\)
Click to expand...

We have: .\(\displaystyle 2^{2x} - (m-2)2^x + 2m+1 \:=\:0\)

Let \(\displaystyle u = 2^x\!:\;\;u^2 - (m-2)u + (2m+1) \:=\:0\)

A quadratic has real roots if its discriminant is positive: .\(\displaystyle b^2-4ac\,>\,0\)

We have: .\(\displaystyle (m-2)^2 - 4(1)(2m+1) \:>\:0\)

. . . . . . . . . \(\displaystyle m^2-4m + 4 - 8m - 4 \:>\: 0\)

. . . . . . . . . . . . . . . . . . .\(\displaystyle m^2-12m \:>\:0\)


\(\displaystyle y \:=\:m(m-12)\) is an up-opening parabola
. . with x-intercepts: \(\displaystyle 0,12\)

It is positive (above the x-axis) "outside" of its x-intercepts.


Therefore: .\(\displaystyle x\,<\,0\,\text{ or }\,x\,>\,12\)

. . . . . . . . \(\displaystyle (\text{-}\infty,0)\:\cup\: (12,\infty)\)
 
soroban said:

We have: .\(\displaystyle 2^{2x} - (m-2)2^x + 2m+1 \:=\:0\)

Let \(\displaystyle u = 2^x\!:\;\;u^2 - (m-2)u + (2m+1) \:=\:0\)

A quadratic has real roots if its discriminant is positive: .\(\displaystyle b^2-4ac\,>\,0\) No, a quadratic equation has real roots if its discriminant is *nonnegative*.

We have: .\(\displaystyle (m-2)^2 - 4(1)(2m+1) \:>\:0\)

. . . . . . . . . \(\displaystyle m^2-4m + 4 - 8m - 4 \:>\: 0\)

. . . . . . . . . . . . . . . . . . .\(\displaystyle m^2-12m \:>\:0\)All of these three inequalities must have \(\displaystyle "\ge" \) symbols.


\(\displaystyle y \:=\:m(m-12)\) is an up-opening parabola
. . with x-intercepts: \(\displaystyle 0,12\)No, it has m-intercepts.

It is positive (above the x-axis) "outside" of its x-intercepts. Again, they are m-intercepts.


Therefore: .\(\displaystyle x\,<\,0\,\text{ or }\,x\,>\,12\)

. . . . . . . . \(\displaystyle (\text{-}\infty,0)\:\cup\: (12,\infty)\)No, we were solving for m-values, not x-values. m = 12 is valid. In addition, the left-hand branch of the inequality is incorrect. For instance, if you check m = -1/4, you will see that y is negative. But it can't be negative, because there are no real solutions to \(\displaystyle 2^x \ = \ \ a \ \ negative \ \ number.\)So, for instance, m = -1/4 is not valid.
Click to expand...
So, the left-hand branch of the inequality is not (-oo, 0). It has to be corrected. The right-hand branch is [12, oo), because m = 12 is also valid.
 
Okay, I tryed this, but none of you (even me) have found the correct asnwer.

Solution = {m e R | m < -1/2 or m >= 12 }

This is the asnwer.
 
danielvnl,


\(\displaystyle y^2 - (m - 2)y + (2m + 1) = 0, \ \ using \ \ 2^x = y. \ \ \)


\(\displaystyle 2^x \ = \ y \ = \ \dfrac{m - 2 \pm\sqrt{m^2 - 12m \ }}{2}. \ \ \ \)


The branch of \(\displaystyle \ m \ge 12 \ \) has already been addressed.


\(\displaystyle 2^x \ \ is \ \ necessarily \ \ positive. \)


Solve:


\(\displaystyle \ \dfrac{m - 2 \pm\sqrt{m^2 - 12m \ }}{2} \ > \ 0 \ \ \implies\)


\(\displaystyle \ m - 2 \pm\sqrt{m^2 - 12m \ } \ > \ 0 \ \ \implies \ \ what \ \ step(s) \ ?\)
 
This is not the way to the answer. I give up.
Unless someone show me the whole development of the question.
I won't do this anymore.

The answer is

Solution = {m e R | m < -1/2 or m >= 12 }

If someone found a way to get this result he did it right, if not, he's wrong.
 
Okay, we have two causes... a) and b)

a) two real positive roots, as you said :


[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]g[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT]


[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]12[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT]


The branch of [FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]≥[/FONT][FONT=MathJax_Main]12[/FONT][FONT=MathJax_Main] [/FONT] has already been addressed.



b) Just ONE positive root, so the PRODUCT of the roots will be negative


gif.latex



simultaneously a) and b)

{m e R | m < -1/2 or m >= 12 }


Is there any objection ?? Am I right ??
 
danielvnl said:
Okay, we have two causes... a) and b)

a) two real positive roots, as you said :


[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]g[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main].[/FONT]


[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]12[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main].[/FONT]


The branch of [FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]≥[/FONT][FONT=MathJax_Main]12[/FONT] has already been addressed.



b) Just ONE positive root, so the PRODUCT of the roots will be negative


gif.latex

I would use implication symbols, not equivalence symbols between each pair of steps.**
simultaneously a) and b)

{m e R | m < -1/2 or m >= 12 }


Is there any objection ?? Am I right ??
Click to expand...
** As in "I have established this result so that I can support the next result."
 
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