solve this sum
A natural number when increased by 84 equals 160 times its reciprocal.Find the number.
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Can u please solve it
Answers
Answered by
124
Hello !
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Let the natural positive number be x
ATQ
x + 84 = 160 ( 1/x)
x² + 84x = 160
x² + 84x - 160 = 0
Here a quadratic equation formed :
Where
a (coefficient of x²) = 1
b (coefficient of x ) = 84
c (constant ) = -160
D (Discriminant ) = b² - 4ac
= 84² - 4 x 1 x 160
= 7056 + 640
= 7696
we are finding roots (zeros) of the quadratic by this quadratic formula
-b+-√D/2a
Two zeros are
160 - √7696 / 2 and 160 + √7696 / 2
________________________________________
Let the natural positive number be x
ATQ
x + 84 = 160 ( 1/x)
x² + 84x = 160
x² + 84x - 160 = 0
Here a quadratic equation formed :
Where
a (coefficient of x²) = 1
b (coefficient of x ) = 84
c (constant ) = -160
D (Discriminant ) = b² - 4ac
= 84² - 4 x 1 x 160
= 7056 + 640
= 7696
we are finding roots (zeros) of the quadratic by this quadratic formula
-b+-√D/2a
Two zeros are
160 - √7696 / 2 and 160 + √7696 / 2
plz , check again
last line is wrong
mistake after D
where ?
last line -b value is wrong
it right because b = -160 so, -b = -(-160)= +160
b =84
the answer should be a natural number, isn't it???
Answered by
2
Answer:
Let the natural positive number be x
ATQ
x + 84 = 160 (1/x)
x² + 84x = 160
x² +84x - 160 = 0
Here a quadratic equation formed:
Where
a (coefficient of x²) = 1
b (coefficient of x ) = 84
c (constant ) = -160
D (Discriminant ) = b² - 4ac
= 84² - 4x1x 160
= 7056 + 640
= 7696
we are finding roots (zeros) of the quadratic by this quadratic formula
-b+-vD/2a
Two zeros are
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