hi, thats where i am stuck..

A sqaure root of a number can be positive or negative. Since we are not given value of X. how can we assume that sqaure root will be positive?

Assume X = -6 . Then numerator is sqaure root of 36. But it will be -6 as -6 * -6 is wat gave us 36 and not 6*6. hence the answer should be 1 always.


Alternative approach. Move the X in the denominator into square root. The question will look like square root of X^2 / X^2. which is equal to square root of 1. which is 1 again.

It would be really helpful if someone can point out the flaw(s) in my reasoning.

sqrt(x^2)= x or -x

Let me explain with ur example

x=-6

sqrt(x^2)/x=sqrt(36)/-6

But sqrt(36)=6 or -6

6/6=1 or -6/6=-1

sqrt(x^2) always has positive and negative root.

You have to take x as positive or negative. You can not take x as positive in numerator and x as negative in denominator.

How about canceling square with square root. That leaves x/x=1
Answer is 1.

I am not taking reverse sign of x in both numerators and denominator.

I am taking same sign of x in both.

However sqrt(x^2)= x or -x

If it is x,sqrt(x^2)/x=x/x=1

If it is -x,sqrt(x^2)/x=-x/x=-1

commonsense27 wrote:You have to take x as positive or negative. You can not take x as positive in numerator and x as negative in denominator.

How about canceling square with square root. That leaves x/x=1
Answer is 1.

Yes by this method answer is always 1. But the correct answer is option E. which is |X|/ X which is still baffling me

Rahul,

Let me explain this in another way.

Assume x=-6

-->sqrt(-6*-6)/-6

Now wat ur assuming is sqrt(-6*-6)=-6.But its sqrt(x^2) sqrt(36)

But note that square root of perfect square has both positive and negative root.

-->sqrt(36)=6 or -6

So the numerator one contains both +ve and -ve sign.For this only,modulus is used.

Hence E.

Let me know if u still have doubts in this.....

selango wrote:Rahul,

Let me explain this in another way.

Assume x=-6

-->sqrt(-6*-6)/-6

Now wat ur assuming is sqrt(-6*-6)=-6.But its sqrt(x^2) sqrt(36)

But note that square root of perfect square has both positive and negative root.

-->sqrt(36)=6 or -6

So the numerator one contains both +ve and -ve sign.For this only,modulus is used.

Hence E.

Let me know if u still have doubts in this.....

Hi Anand,

I understand the flaw in my reasoning but wat about the substitution of denominator in the square root?

Rahul,

Tats only am saying sqrt(x^2)=x or -x

you are taking sqrt(-6*-6)/-6=-6/-6=1.We can't substitute in this way.

sqrt(-6*-6)=sqrt(36)=sqrt(6*6)

you should substitute x as sqrt(x^2) not as sqrt(x*x)

Got it?

selango wrote:Rahul,

Tats only am saying sqrt(x^2)=x or -x

you are taking sqrt(-6*-6)/-6=-6/-6=1.We can't substitute in this way.

sqrt(-6*-6)=sqrt(36)=sqrt(6*6)

you should substitute x as sqrt(x^2) not as sqrt(x*x)

Got it?

Hi Anand,

sorry to irritate u...I understand ur logic ...and it works well for my approach 1 where i am making the mistake of taking the negative sign.

But according to the second approach ..the question can be restated as sqrt(x^2/x^2). in this case wont the answer be 1 in all cases? I hope I am not testing ur patience

Interesting..I didn't see this approach..

Ok let substitute the same value x=-6 in the equation.

sqrt(-6*-6)/-6=sqrt(36)/-6

sqrt(36)/sqrt(36)[Here you cancel both sides]

we know sqrt(36)=6 or -6

-->sqrt(36)/sqrt(36)=6/6 or -6/-6 or 6/-6 or -6/6

result is either 1 or -1

Always remember square root of perfect square has both +ve and -ve roots.

Hope this clarify:)

selango wrote:Interesting..I didn't see this approach..

Ok let substitute the same value x=-6 in the equation.

sqrt(-6*-6)/-6=sqrt(36)/-6

sqrt(36)/sqrt(36)[Here you cancel both sides]

we know sqrt(36)=6 or -6

-->sqrt(36)/sqrt(36)=6/6 or -6/-6 or 6/-6 or -6/6

result is either 1 or -1

Always remember square root of perfect square has both +ve and -ve roots.

Hope this clarify:)

Hi Anand,

great explanation and thanks a ton for ur time .

Glad u got it:)

sqrt(x^2)/x=((x^2)^1/2)/x =x^1-1=x^0=1

It does not matter what is value of x is. Answer is 1. If book is saying otherwise, then book needs to be corrected.

BTW, What is the source of this problem?