Question

x-1/x=5 find the value of x^2+1/x2
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Answer 1

Solution:-

Given

$\rm \to \: x - \dfrac{1}{x} = 5$

To find the value of

$\rm \to \: {x}^{2} - \dfrac{1}{ {x}^{2} } = 5$

So now take

$\rm \to \: x - \dfrac{1}{x} = 5$

Squaring on both side

$\rm \to \bigg(x - \dfrac{1}{x} \bigg) ^{2} = (5) ^{2}$

Using this identity

$\rm \to(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

We get

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x} = 25$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times \cancel{x} \times \dfrac{1}{ \cancel{x}} = 25$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 25$

$\rm \to {x}^{2} + \dfrac{1}{ {x}^{2} } = 25 + 2$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 27$

So value of

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: \: is \: \: 27$

Answer is 27

Answer 2

Step-by-step explanation:

Required answer:-

$x-{\dfrac {1}{x}}=5$

• Now use identity

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

• Take squares of both sides

${:}\mapsto$${(x-{\dfrac {1}{x}})^{2}={5}^{2}}$

${:}\mapsto$${x}^{2}+{\dfrac {1}{x}}{}^{2}-2×{\cancel {x}}×{\dfrac {1}{{\cancel{x}}}}=25$

${:}\mapsto$${x}^{2}+{\dfrac {1}{{x}^{2}}}-2=25$

• Now interchange both sides

${:}\mapsto$${x}^{2}+{\dfrac {1}{{x}^{2}}}=25+2$

${:}\mapsto$${\underline {\boxed {\bf {{x}^{2}+{\dfrac {1}{{x}^{2}}}=27}}}}$