Question

The number of irrational roots of 2x4 + x3 – 11x2 + x + 2 = 0 is
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Answer 1

SOLUTION

TO DETERMINE

The number of irrational roots of

$\displaystyle \sf{2 {x}^{4} + {x}^{3} - 11 {x}^{2} + x + 2 = 0}$

EVALUATION

Here the given equation is

$\displaystyle \sf{2 {x}^{4} + {x}^{3} - 11 {x}^{2} + x + 2 = 0}$

We find the roots as below

$\displaystyle \sf{2 {x}^{4} + {x}^{3} - 11 {x}^{2} + x + 2 = 0}$

Dividing both sides by x² we get

$\displaystyle \sf{ \implies \: 2 {x}^{2} + x- 11 + \frac{1}{x} + \frac{2}{ {x}^{2} } = 0}$

$\displaystyle \sf{ \implies \: 2 \bigg({x}^{2} + \frac{1}{ {x}^{2} } \bigg) + \bigg( x + \frac{1}{x} \bigg) - 11 = 0}$

$\displaystyle \sf{ \implies \: 2 {\bigg({x}^{} + \frac{1}{ {x}^{} } \bigg) }^{2} - 4 + \bigg( x + \frac{1}{x} \bigg) - 11 = 0}$

$\displaystyle \sf{ \implies \: 2 {\bigg({x}^{} + \frac{1}{ {x}^{} } \bigg) }^{2} + \bigg( x + \frac{1}{x} \bigg) - 15 = 0}$

$\displaystyle \sf{ Let \: \: y \: = \bigg( x + \frac{1}{x} \bigg)}$

Then from above we get

$\displaystyle \sf{ \implies \: 2 {y}^{2} + y - 15 = 0}$

$\displaystyle \sf{ \implies \: y = - 3 \: \: \: and \: \: \frac{5}{2} }$

Now y = - 3 gives

$\displaystyle \sf{ \bigg( x + \frac{1}{x} \bigg) = - 3}$

$\displaystyle \sf{ \implies \: {x}^{2} + 3x + 1 = 0 }$

$\displaystyle \sf{ \implies \: x = \frac{ - 3 \pm \: \sqrt{5} }{2} }$

$\displaystyle \sf{Again \: \: \: y = \frac{5}{2} \: \: gives }$

$\displaystyle \sf{ \bigg( x + \frac{1}{x} \bigg) = \frac{5}{2} }$

$\displaystyle \sf{ \implies \: 2 {x}^{2} - 5x + 2 = 0 }$

$\displaystyle \sf{ \implies \: x = \frac{1}{2} \: \: \: and \: \: 2 }$

So all four roots are

$\displaystyle \sf{ \frac{ - 3 + \: \sqrt{5} }{2} \: , \: \frac{ - 3 - \: \sqrt{5} }{2} \: , \: \frac{1}{2} \: , \: 2 }$

So there are two irrational roots

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