Find the value of x squared + 1/x squared if x + 1/x = 5

hard CBSE JEE-MAIN 2 min read

Question

If x+1x=5x + \dfrac{1}{x} = 5, find the value of x2+1x2x^2 + \dfrac{1}{x^2}.

Solution — Step by Step

We need to go from x+1xx + \frac{1}{x} to x2+1x2x^2 + \frac{1}{x^2}. The connection is through squaring.

The key identity: (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Here, let a=xa = x and b=1xb = \frac{1}{x}. Then ab=x1x=1ab = x \cdot \frac{1}{x} = 1.

(x+1x)2=52=25\left(x + \frac{1}{x}\right)^2 = 5^2 = 25

Expanding the left side:

x2+2x1x+1x2=25x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 25 x2+2+1x2=25x^2 + 2 + \frac{1}{x^2} = 25
x2+1x2=252=23x^2 + \frac{1}{x^2} = 25 - 2 = \boxed{23}

Why This Works

This is a classic “reciprocal expression” problem. The trick is recognizing that squaring x+1xx + \frac{1}{x} naturally produces x2+1x2x^2 + \frac{1}{x^2} with an extra constant (+2+2) that we can subtract away.

The reason the middle term simplifies so cleanly is that x1x=1x \cdot \frac{1}{x} = 1 regardless of the value of xx. This “reciprocal property” makes these problems very tractable.

Alternative Method — Numerical Verification

We can find xx from x+1x=5x + \frac{1}{x} = 5: multiply by xx to get x25x+1=0x^2 - 5x + 1 = 0.

x=5±212x = \frac{5 \pm \sqrt{21}}{2}

Take x=5+2125+4.5824.79x = \frac{5 + \sqrt{21}}{2} \approx \frac{5 + 4.58}{2} \approx 4.79

Then x222.94x^2 \approx 22.94 and 1x20.044\frac{1}{x^2} \approx 0.044.

x2+1x222.94+0.04423x^2 + \frac{1}{x^2} \approx 22.94 + 0.044 \approx 23

The algebraic method is far faster than computing irrational roots.

Common Mistake

The most common error: squaring x+1xx + \frac{1}{x} to get x2+1x2x^2 + \frac{1}{x^2} directly, forgetting the middle term +2+2. (a+b)2a2+b2(a+b)^2 \neq a^2 + b^2 — the cross term 2ab2ab must always be included. Here, students write 52=25=x2+1x25^2 = 25 = x^2 + \frac{1}{x^2} and get 25 instead of 23.

This pattern extends: if you know x2+1x2x^2 + \frac{1}{x^2}, you can find x4+1x4x^4 + \frac{1}{x^4} by squaring again: (x2+1x2)2=x4+2+1x4\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + 2 + \frac{1}{x^4}, so x4+1x4=2322=527x^4 + \frac{1}{x^4} = 23^2 - 2 = 527.

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