Question

A positive number when decreased by 4 is equal to 32 times of the reciprocal of the number. Find the number.
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Answer 1

Answer:

Let N be the number in question.

Then N-4 = 21×(1/n).

So N-4 = 21/n.

So N-4-21/N = 0.

So N(N-4-21/N) = N×0.

So N^2–4N-21 = 0.

This is the quadratic equation aN^2+bN+C = 0 in N, with a = 1, b = -4, and c = -21.

So by the quadratic equation:

N = [(-b)+/-√(b^2-4ac)]/(2a)

= {[-(-4)]+/-√[(-4)^2-4×1×(-21)]}/(2×1)

= {4+/-√[16-(-84)]}/2

= [4+/-√(16+84)]/2

= (4+/-√100)/2

= (4+/-10)/2

= -6/2 or 14/2

= -3 or 7.

So N = -3 or N =7.

CHECK:

If N = -3, then N^2-4N-21 = (-3)^2-4×(-3)-21 = 9-(-12)-21 = 9+12-21 = 0. ✓

If N = 7, then N^2-4N-21 = 7^2-4×7-21 = 49-28-21 = 0. ✓

Since N is positive, N = 7.

CHECK:

N-4 = 7-4 = 3.

and 21×(1/N) =21/N = 21/7 = 3.

So N-4 = 21×(1/N). ✓

So the answer is 7.

Answer 2

Required Answer :-

Given :-

• Positive number when decreased by 4
• And it is equal to equal to 32 times of the reciprocal of the number

To find :-

• A positive number

Solution :-

Now,

Let the number be x .

And according to the question ,

$:\implies\sf x - 4 = \dfrac{32}{x}$

$:\implies\sf (x - 4 )x= 32$

$:\implies\sf {x}^{2} - 4 x= 32$

$:\implies\sf {x}^{2} - 4 x - 32 = 0$

Now we have to factorise it

$:\implies\sf {x}^{2} - 4 x - 32 = 0$

$:\implies\sf {x}^{2} + 4 x - 8x - 32 = 0$

$:\implies\sf {x}(x + 4 ) - 8(x + 4) = 0$

$:\implies\sf (x- 8)(x + 4) = 0$

Now , x = 8 and x = -4

But , the number cannot be negetive . therefore,

x = 8

Check :-

$:\implies\sf x - 4 = \dfrac{32}{x}$

★ x = 8

LHS ,

$:\implies\sf 8 - 4 = 4$

RHS ,

$:\implies\sf \dfrac{32}{8} =4$

hence , LHS = RHS

hence , The positive number is 8