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Question : 24 of 39

Marks: +1, -0

Solving equation

\left(x-\frac{1}{x}\right)^{2}-6\left(x+\frac{1}{x}\right)+12=0

we get roots as follows

0
1
$-1$
None

Solution:

Concept:

Rewrite

\left(x-\frac{1}{x}\right)^2

in terms of

x+\frac{1}{x}

using the identity

\left(x+\frac{1}{x}\right)^2 = \left(x-\frac{1}{x}\right)^2 + 4

.

Explanation:

Let

t = x + \frac{1}{x}

.

Then

\left(x-\frac{1}{x}\right)^2 = \left(x+\frac{1}{x}\right)^2 - 4 = t^2 - 4

.

Substitute into the equation:

(t^2-4) - 6t + 12 = 0 \Rightarrow t^2 - 6t + 8 = 0

.

Factor:

(t-2)(t-4)=0

, so

t=2

or

t=4

.

Case

t=2

:

x + \frac{1}{x}=2 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2=0 \Rightarrow x=1

.

Case

t=4

:

x + \frac{1}{x}=4 \Rightarrow x^2 - 4x + 1 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}

.

Thus the roots are

1

,

2+\sqrt{3}

,

2-\sqrt{3}

.

Answer:

Among the given options,

1

is a root.

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