Question

Solve the equation :

$\bf \: x + \frac{1}{x} = 2 \times \frac{1}{20} , x ≠ 0$

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Answer 1

Given :

• $\sf\:x+\frac{1}{x}=2\times\frac{1}{20}$

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To find :

• The value of x.

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Concept :

Simply we would solve these question following simple calculation. However if we get quadratic equation we would solve it by breaking middle term if it's not possible we would use the quadratic formula as follows -

$\sf\:\maltese$ $\sf\:\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

Doing so we would get the value or root of the equation.

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Solution :

$\sf\:x+\frac{1}{x}=2\times\frac{1}{20}$

$\sf\rightarrow\:x+\frac{1}{x}=\frac{2}{20}$

$\sf\rightarrow\:\frac{x^{2}+1}{x}=\frac{2}{20}$

$\sf\rightarrow\:20(x^{2}+1)=2x$

$\sf\rightarrow\:20x^{2}+20=2x$

$\sf\rightarrow\:20x^{2}-2x+20=0$

$\sf\rightarrow\:2(10x^{2}-x+10)=0$

$\sf\rightarrow\:10x^{2}-x+10=0$

Using quadratic formula,

$\sf\:\dag$ $\sf\:\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

Here in our expression,

⠀⠀⠀⠀⠀⠀⠀ a = 10 , b = -1 and c = 10

$\sf\to\:\frac{-(-1)±\sqrt{(-1)^{2}-4 \times 10 \times 10}}{2 \times 10}$

$\sf\to\:\frac{1±\sqrt{1-4 \times 100}}{20}$

$\sf\to\:\frac{1±\sqrt{1-400}}{20}$

$\sf\to\:\frac{1±\sqrt{-399}}{20}$

$\bf\therefore\:Either$

⠀⠀⠀⠀⠀⠀❒ $\sf\:\frac{1+\sqrt{-399}}{20}$

$\bf\:oR$

⠀⠀⠀⠀⠀⠀❒ $\sf\:\frac{1-\sqrt{-399}}{20}$

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Note :

The square root of a negative number is not a real number hence discriminant is -399.

Answer 2

Answer:

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