Question

the denominator of the fraction is 1 more than twice its numerator if the sum of fraction and its reciprocal is 58 by 21 find the fraction

Answer 1

Answer:the question is wrong .you cannot solve the quadratic equation see attachment

Step-by-step explanation:

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption,

Numerator of fraction be p

So,

Denominator of fraction = (2p + 1)

Also,

Fraction = p/(2p + 1)

Reciprocal = (2p + 1)/p

Situation,

⇒ p/(2p + 1) + (2p + 1)/p = 58/21

⇒ 21[p² + (2p + 1)²] = 58p(2p + 1)

⇒ 21[5p² + 4p + 1] = 116p² + 58p

⇒ 11p² - 26p - 21 = 0

⇒ 11p² - 33p + 7p - 21 = 0

⇒ 11p(p - 3) + 7(p - 3) = 0

⇒ (p - 3)(11p + 7) = 0

⇒  p - 3 = 0

⇒ p = 3

Or,

⇒ 11p + 7 = 0

⇒ p = -7/11 (This is not applicable)

Hence,

⇒ p = 3

Numerator = p = 3

Denominator

= 2p + 1

= 2(3) + 1

= 6 + 1

= 7

So,

Fraction = 3/7

Therefore,

Required fraction = 3/7