Answers

2015-11-26T17:35:26+05:30
Given 
[(x-2)/(x-1)]×[(x+4)/(2x+2)] = 0
[(x-2)(x+4)] / [(x-1)(2x+2)] = 0
(x² + 4x - 2x - 8)/ (2x² + 2x - 2x -2) = 0
(x² + 2x - 8) / (2x² - 2) = 0
x² + 2x - 8 = 0 × (2x² - 2)
⇒x² + 2x - 8 = 0 ------------------- eq (1)
 Standard eq of Q.E is ax² + bx +c = 0  --------- (2)
Comparing (1) and (2)  we get
 a = 1 , b = 2 and c= -8
With the help of discriminant  we can find the nature of their roots.
Discriminant(Δ) = b² - 4ac
2² - 4(1)(-8)  =  4+32  =  36
If b² - 4ac is greater than 0 then the roots are REAL AND DISTINCT
∴The nature of the roots  are real and distinct.


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