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Number of irrational roots of the equation $(x^2-3x+1)(x^2+3x+2)(x^2-9x+20)=-30$ are

$\begin{array}{1 1}(A)\;0&(B)\;2\\(C)\;4&(D)\;6\end{array} $

Can you answer this question?
 
 

1 Answer

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$x^2+3x+2=(x+1)(x+2)$
$x^2-9x+20=(x-4)(x-5)$
$(x^2+3x+2)(x^2-9x+20)=(x+1)(x-4)(x-2)(x-5)$
$(x^2+3x+2)(x^2-9x+20)=(x^2-3x-4)(x^2-3x-10)$
Now putting $x^2-3x=y$,above equation can be written as,
$(y+1)(y-4)(y-10)=-30$
Roots of this equation are $5,4,\pm\sqrt{30}$
Use these to obtain roots of x
Hence (D) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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