The difference of two natural numbers is 3 and the difference of their reciprocals is 3 / 28. Find the numbers.
Quadratic Equations
Class 10
Answers
Answered by
2
so let the larger one be X and another be y
x-y=3
X=3+y
now
1/X -1/y = 3/28
1/3+y -1/y =3/28
y-3-y/y^2+3y=3/28
3y^2+9y+84=0
y^2+3y+28=0
y^2+7y-4y+28=0
y(y+7)-4(y+7)=0
(y+7)(y-4)=0
y=-7 or 4
negative value discard as natural no. is given
so smaller no. is 4
and larger no. is 3+4=7
x-y=3
X=3+y
now
1/X -1/y = 3/28
1/3+y -1/y =3/28
y-3-y/y^2+3y=3/28
3y^2+9y+84=0
y^2+3y+28=0
y^2+7y-4y+28=0
y(y+7)-4(y+7)=0
(y+7)(y-4)=0
y=-7 or 4
negative value discard as natural no. is given
so smaller no. is 4
and larger no. is 3+4=7
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Answered by
5
Let the two natural numbers be x and y.
(i)
Given that difference of two natural numbers is 3.
⇒ x - y = 3.
⇒ x = y + 3.
(ii)
Given that difference of their reciprocals is 3/28.
⇒ (1/y) - (1/x) = 3/28
⇒ (1/y) - (1/y + 3) = (3/28)
⇒ 84 = 3y^2 + 9y
⇒ 3y^2 + 9y - 84 = 0
⇒ y^2 + 3y - 28 = 0
⇒ y^2 +7y - 4y - 28 = 0
⇒ y(y + 7) - 4(y + 7) = 0
⇒ (y + 7)(y - 4) = 0
⇒ y = 4,-7{Neglect -ve values}
⇒ y = 4.
Substitute y = 4 in (i), we get
⇒ x = y + 3
⇒ x = 7.
Therefore, the required natural numbers are 7 and 4.
Hope this helps!
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