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find the values of b for which the equation 2log 1/25 (bx+28) = -log 5 (12-4x-x 2 ) has only one solution.

find the values of b for which the equation 2log1/25(bx+28) = -log5(12-4x-x2) has only one solution.

Grade:11

3 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

2log1/25(bx+28) = -log5(12-4x-x2), log(1/25)(bx+28)2=log5(12-4x-x2)-1

(12-4x-x2)-1=bx+28, (bx+28)(12-4x-x2)=1

 

Swapnil Saxena
102 Points
9 years ago

= 2 log1/25 (bx+28) = -log5(12-4x-x2)

= -2 log25 (bx+28) = -log5(12-4x-x2)

= - log5(bx+28) = -log5(12-4x-x2)

= log5(bx+28) = log5(12-4x-x2)

= bx+28 = 12-4x-x2

= x2+ (4+b)x +16

If the quad has only one sol. then D=0

(4+b)2-4(16) =  b2 + 8b - 48=0

(b+12)(b-4)=0

b=12,4


saket shrivastava
36 Points
9 years ago

method as above ans is -12, 4 for x belongs (-6,2)

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