If twice the reciprocal ofx-1=one sixth of x find both roots................next
find 2 consecutive even numbes in such a way tha the sum of squares of these would be 52..nxt
if the sum of a number and its reciprocal it 26/5 gind the nubers

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2014-11-29T19:35:12+05:30

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So the reciprocal of x - 1 is 1/(x - 1) 

so according to question 2/(x - 1) = x/6
                                   ⇒ 12 = x(x - 1)
                                    ⇒ x² - x - 12 = 0
                                   ⇒ x² - 4x + 3x - 12 = 0
                                   ⇒ x(x - 4) + 3(x - 4) = 0
                                   ⇒ (x + 3)(x - 4) = 0

so x = -3 or 4

2) so let the two consecutive even numbers be x, x+2

 so according to question x² + (x + 2)² = 52
⇒ 2x² + 4x - 48 = 0 
⇒x² + 2x - 24 = 0
⇒ x² + 6x - 4x - 24  = 0
⇒ x(x + 6) - 4(x + 6) = 0
⇒(x - 4)(x + 6) = 0

so the even numbers are 2,6 or -6, -4

3) let the number be x 

so x + 1/x = 26/5

⇒ 5x² - 26x + 5 = 0
⇒ 5x² - 25x - x + 5 = 0
⇒ 5x(x - 5) - 1(x - 5) = 0
⇒(5x - 1)(x - 5) = 0

so x = 1/5 or 5  

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