How many real number a<-1 satisfies int_0^1 [ 1 / (x^2 + a) ] = ln(1/2) / (2 sqrt{-a})

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Violette

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How many real number a<-1 satisfies the following?
[math]∫_0^1\dfrac{1}{x^2+a}=\dfrac{1}{2\sqrt{-a}}ln\left(\dfrac{1}{2}\right)?[/math]I've tried subtitute [imath]a=-b[/imath]
[math]∫_0^1\dfrac{1}{(x-\sqrt{b})(x+\sqrt{b})}dx=\dfrac{1}{2\sqrt{b}}(ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|-ln|-1|) =\dfrac{1}{2\sqrt{-a}}ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|[/math][math]=>\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|=\frac{1}{2}<=>b= \frac{1}{9} \text{ or } b = 9[/math][math]a=-b\text{ and } a<-1=>b>1 =>b=9<=>a=-9[/math]Was this right ? I posted this because I want to ask for the complex number method or better method^^
 
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Violette said:
[imath]∫_0^1\dfrac{1}{x^2+a}=\dfrac{1}{2\sqrt{-a}}ln(\dfrac{1}{2})?[/imath]
I've tried subtitute [imath]a=-b[/imath]
[math]∫_0^1\dfrac{1}{(x-\sqrt{b})(x+\sqrt{b})}dx=\dfrac{1}{2\sqrt{b}}(ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|-ln|-1|) =\dfrac{1}{2\sqrt{-a}}ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|[/math][math]=>\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|=\frac{1}{2}<=>b= \frac{1}{9} \text{ or } b = 9[/math][math]a=-b\text{ and } a<-1=>b>1 =>b=9<=>a=-9[/math]Was this right ? I posted this because I want to ask for the complex number method or better method^^
Click to expand...

Please share the intermediate steps.
 
Violette said:
[imath]∫_0^1\dfrac{1}{x^2+a}=\dfrac{1}{2\sqrt{-a}}ln(\dfrac{1}{2})?[/imath]
I've tried subtitute [imath]a=-b[/imath]
[math]∫_0^1\dfrac{1}{(x-\sqrt{b})(x+\sqrt{b})}dx=\dfrac{1}{2\sqrt{b}}(ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|-ln|-1|) =\dfrac{1}{2\sqrt{-a}}ln\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|[/math][math]=>\left|\dfrac{1-\sqrt{b}}{1+\sqrt{b}}\right|=\frac{1}{2}<=>b= \frac{1}{9} \text{ or } b = 9[/math][math]a=-b\text{ and } a<-1=>b>1 =>b=9<=>a=-9[/math]Was this right ? I posted this because I want to ask for the complex number method or better method^^
Click to expand...
Check your answer by plugging a=-9 into the problem and carrying out the definite integral.

You mention complex numbers; what is the context of the problem? That is, what course are you taking (and what is the current topic), or where else does it come from?
 
x^2 + a = x^2 + (sqrt(a))^2. You can't factor the sum of squares.
If you replaced a with -b, then x^2 + (-b) = x^2 + sqrt(-b))^2 which is still a sum of squares.
What you did was: x^2 + a = x^2 + (-b) = x^2 - b = x^2 - [sqrt(b)]^2. Can you find your error?
If you could factor the sum of squares you'd have learned how to do this in algebra

Example: 10^2 + 9 = 10^2 - (-9) = 10^2 - ?^2
 
Steven G said:
If you replaced a with -b, then x^2 + (-b) = x^2 + sqrt(-b))^2 which is still a sum of squares.
Click to expand...
Ummm ... no. [imath](\sqrt{-b})^2\ne-b[/imath] if [imath]b=-a>1[/imath].

The problem is deeper than that. You can factor [imath]x^2-b[/imath].

If [imath]a = -9[/imath], then [imath]x^2+a=x^2-9[/imath] which is a difference of squares.

And in fact, the check I suggested works. What we'd like to see is the rest of the work (some substitutions one might have used would be inappropriate). And, why complex numbers are relevant.

Possibly some of us are suffering from didn't-read-the-title syndrome:

How many real number a<-1 satisfies the following?​

 
khansaheb said:
Please share the intermediate steps.
Click to expand...
[math]2|1-\sqrt{b}|=|1+\sqrt{b}|<=>4(1-\sqrt{b})^2=(1+\sqrt{b})^2<=>4(1-2\sqrt{b}+b)=1+2\sqrt{b}+b <=>3(\sqrt{b})^2-10\sqrt{b}+3=0<=>\sqrt{b}=\frac{1}{3} \text{ or } \sqrt{b}=3<=>b=\frac{1}{9} \text { or } b=9[/math]
 
Dr.Peterson said:
Check your answer by plugging a=-9 into the problem and carrying out the definite integral.

You mention complex numbers; what is the context of the problem? That is, what course are you taking (and what is the current topic), or where else does it come from?
Click to expand...
it's a problem from a book of 20 tests illustrated the 2023 university entrance exam, i think the question's topic is integral
 
Dr.Peterson said:
Ummm ... no. [imath](\sqrt{-b})^2\ne-b[/imath] if [imath]b=-a>1[/imath].

The problem is deeper than that. You can factor [imath]x^2-b[/imath].

If [imath]a = -9[/imath], then [imath]x^2+a=x^2-9[/imath] which is a difference of squares.

And in fact, the check I suggested works. What we'd like to see is the rest of the work (some substitutions one might have used would be inappropriate). And, why complex numbers are relevant.

Possibly some of us are suffering from didn't-read-the-title syndrome:

How many real number a<-1 satisfies the following?​

Click to expand...
instead of plugging [math]a=-b \text { I used} (x-\sqrt{a}i)(x+\sqrt{a}i) [/math]^^
I think they are the same
 
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