If x^2 + 9/x^2 = 31, what is the value of x - 3/x?

option D.

To find : x-3/x. Let it be t.
=> x-3/x = t
=> (x^2 + 9/x^2) - 2*x*3/x = t^2 (Squaring both sides).
=> (31) - 2*3 = 25
=> t^2 = 25. Thus t=5 or t=-5.

=> option D.

Dont get this part

- 2*x*3/x

Have just used the formula (a-b)^2 = a^2 + b^2 - 2*a*b.
Here a is x and b is 3/x.

Consider Kudos if this helps.

Hi ritikk,

if x-3/x is squared then resultant equation after substitution will be 31 - (6/x). How do you arrive at 5 from this when x is still present?

very nice question but needs smart eyes to catch the idea.
Are there any alternate approach ?

Bunuel thanks a lot. The misunderstanding is due to the font type not sure if it was x-3 the whole term by x or just x minus 3/x. That issue with members posting can solve the issue i think any way out or any suggestion like you have typed now appropriately? Bunuel

Mathematically x - 3/x can only mean x minus 3/x. If it were x-3 over x it would e written as (x - 3)/x.

Let’s square x - 3/x first:

(x - 3/x)^2 = x^2 - 6 + 9/x^2

Since x^2 + 9/x^2 = 31, we have x^2 - 6 + 9/x^2 = 31 - 6 = 25.

So, (x - 3/x)^2 = 25 and:

x - 3/x = ±5

We see that only 5 is among the answer choices.

Answer: D

Hey pushpitkc,

can you shed some light on the solution above ..i dont get the steps (highlighted) after this "Since x^2 + 9/x^2 = 31"

many thanks

dave13

You can convert any a^2+b^2 to a^2+b^2-2ab+2ab=(a-b)^2+2ab
x^4 + 9/x^2 = x^4 + 9/x^2 - 2*(x^2 *3/x^2) + 2*(x^2 *3/x^2)

Hey dave13

We are given \(x^2 + \frac{9}{x^2} = 31\) in the queston stem -> Lets call this equation (1)

Once, we have squared the equation, \((x - \frac{3}{x})^2\) to get \(x^2 - 6 + \frac{9}{x^2}\)

After re-arranging \((x - \frac{3}{x})^2 = x^2 + \frac{9}{x^2} - 6 = 31 - 6 = 25\) [from equation (1)]

Now, \((x - \frac{3}{x})^2 = 25\) -> \((x - \frac{3}{x}) = \sqrt{25} = +5\) or \(-5\)

Therefore, the answer option which matches the value of \((x - \frac{3}{x})\) is 5(Option D)

Hope this helps you

One could try and substitute values.

if x=1 the expression (x^2+9/x^2) = 10
if x=2, 4+(9/4)=6.25
if x=3, 10
if x=4, 16&9/16
if x=5, 25&9/25
if x=6 36&1/4

We want the expression to =31 so x=5.5 is a good estimate.

x-3/x=

5.5-(3/5.5)=

5.5 - (little more than .5)=close to 5

Let \(a\) = the correct answer.

\(x - \frac{3}{x}=a\)

Since \((m-n)^2 = m^2 + n^2 - 2mn\), squaring both sides yields the following:
\((x - \frac{3}{x})^2=a^2\)

\(x^2 - \frac{9}{x^2} - 2(x)(\frac{3}{x})=a^2\)

\(x^2 - \frac{9}{x^2} - 6=a^2\)

Substituting \(x^2 + \frac{9}{x^2} = 31\) into the resulting blue equation, we get:
\(31 - 6=a^2\)

\(25 = a^2\)

±\(5=a\)

let y= 3/x => xy=3 ----- 1
x^2+y^2= 31 --- 2
x - y = ?

(x-y)^2 = x^2 + y^2- 2xy ---- 3
subs 1 and 2 in 3
we get
(x-y)^2 = 31-6 = 25 = 5^2

so x-y = 5

Simple logic and a basic understanding of Algebraic identities will tell you that x – \(\frac{3}{x }\)cannot be as big a number as 36.

If x – \(\frac{3}{x}\) = 36, \(x^2\) + \(\frac{9}{x^2}\) will not be as small as 31 since squaring 36 will give us a large number like 1296.

What do I mean?

\((x – \frac{3}{x}) ^2\) = \(x^2 + \frac{9}{x^2} – 2*x * \frac{3}{x}\)

Since (x-\(\frac{3}{x}\)) = 36, \((36)^2\) = \(x^2\) + \(\frac{9 }{ x^2}\) – 6 and therefore,\(x^2\) + \(\frac{9}{x^2}\) cannot be as small as 31.

Answer options A and B can be easily eliminated. A little more deliberation will tell you that 9 also is big enough to be eliminated, so answer option C also can be ruled out.

On the other hand, 3 is too small a value for x – \(\frac{3}{x}\) which helps us eliminate answer option E.

The correct answer option is D.

If (x-\(\frac{3}{x}\)) = 5, \((x-\frac{3}{x})^2\) = 25.

Therefore,\( x^2 + \frac{9}{x^2}\) – 6 = 25 or \(x^2 + \frac{9}{x^2}\) = 31, as given in the question data.

Hope that helps!
Aravind B T

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