If the sum of the roots of the equation 1 x + a + 1 x + b = 1 c is zero, prove that the product of the roots is - 1 2 ( a 2 + b 2 ) .

Given, 1 x + a + 1 x + b = 1 c Take the L.C.M of denominator x + b + x + a x + a x + b = 1 c ⇒ 2 x + b + a ( x + a ) ( x + b ) = 1 c ⇒ 2 x c + b c + a c = x 2 + x b + a x + a b ⇒ x 2 + x ( b + a - 2 c ) + a b - c ( b + a ) = 0 Now , assume that roots of the quadratic equation is 0 So, b + a - 2 c = 0 ⇒ b + a = 2 c and a b - c ( b + a ) ⇒ a b - ( a + b ) ( a + b ) 2 ⇒ 2 a b - ( a 2 + b 2 + 2 a b ) 2 ⇒ - 1 2 ( a 2 + b 2 ) Hence, proved

The sum of the number and its reciprocal is 37 6 . Find the number.

Let the first number be x so the reciprocal of the number is 1 x According to the question, x + 1 x = 37 6 ⇒ x 2 + 1 x = 37 6 ⇒ x 2 + 1 = 37 x 6 ⇒ 6 x 2 + 6 = 37 x [Multiplying both the sides by 6 ] ⇒ 6 x 2 - 37 x + 6 By using quadratic formula, we get x = - - 37 ± - 37 2 - 4 × 6 × 6 2 × 6 ⇒ x = 37 ± 1369 - 144 12 ⇒ x = 37 ± 35 12 So, x = 37 + 35 12 = 6 Or, x = 37 - 35 12 = 1 6 So, the number is 6 or 1 6 .

The sum of two numbers is 7 and the difference of their reciprocal is 1 12 . What are the two numbers?

Let the first number be x so the other number is 7 - x According to the question, 1 x - 1 7 - x = 1 12 ⇒ 7 - x - x x 7 - x = 1 12 ⇒ 84 - 24 x = 7 x - x 2 ⇒ x 2 - 7 x - 24 x + 84 = 0 ⇒ x 2 - 31 x + 84 = 0 By using quadratic formula, we get x = - - 31 ± - 31 2 - 4 × 1 × 84 2 × 1 ⇒ x = 31 ± 961 - 336 2 ⇒ x = 31 ± 25 2 ⇒ x = 31 + 25 2 = 28 Or, x = 31 - 25 2 = 3 So, if one number is 28 , then the other number is 7 - 28 = - 21 . Or, if one number is 3 , then the other number is 7 - 3 = 4 .