Question

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Answer 1

We have ,

$x + \frac{1}{x} = 7$

Using

【(a+b)^2 = a^2 + b^2 + 2ab】

$(x + \frac{1}{x} ) ^{2} = x ^{2} + ( \frac{1}{x} )^{2} + 2.x. \frac{1}{x}$

$7 ^{2} = x^{2} + \frac{1}{x ^{2} } + 2$

$49 = x^{2} + \frac{1}{x^{2} } + 2$

$49 - 2 = x ^{2} + \frac{1}{x ^{2} }$

$x ^{2} + \frac{1}{x ^{2} } = 47$

Answer 2

Heya ☺

Given that ,

$x + \frac{1}{x} = 7$

using identity ,

$(a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab$

we get ,

$(x + \frac{1}{x} ) {}^{2} = (7) { }^{2} \\ x { }^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} = 49 \\ x {}^{2} + \frac{1}{ {x {}^{2} }^{} } + 2 = 49 \\ x {}^{2} + \frac{1}{x {}^{2} } = 49 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 47$

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