The number obtained by adding the squares of a number and
its reciprocal is greater by 7/4 than the sum of the number and
4
its reciprocal. Find the number.
Answers
Solution :-
Let the number be x
Reciprocal = 1/x
Given :
Sum of square of number and square of its reciprocal = 7/4 greater than sum of number and it's reciprocal
⇒ x² + 1/x² = x + 1/x + 7/4
Adding 2 on both sides
⇒ x² + 1/x² + 2 = x + 1/x + 7/4 + 2
It can be written as
⇒ ( x )² + ( 1/x )² + 2( x )( 1/x ) = x + 1/x + 15/4
⇒ (x + 1/x)² = x + 1/x + 15/4
[ Because (a + b) ² = a² + b² + 2ab ]
Substituting p = x + 1/x in the given equation
⇒ p² = p + 15/4
⇒ p² - p = 15/4
⇒ 4p² - 4p = 15
⇒ 4p² - 4p - 15 = 0
Splitting the middle term
⇒ 4p² - 10p + 6p - 15 = 0
⇒ 2p(2p - 5) + 3(2p - 5) = 0
⇒ (2p + 3)(2p - 5) = 0
⇒ 2p + 3 = 0 or 2p - 5 = 0
I) When 2p + 3 = 0
⇒ 2(x + 1/x) + 3 = 0
⇒ 2x + 2/x + 3 = 0
⇒ 2x² + 3x + 2 = 0
Here, Discriminant = b² - 4ac = 3² - 4(2)(2) = 9 - 16 = - 7 which is less than 0
Therefore the equation has no real roots
II) When 2p - 5 = 0
⇒ 2(x + 1/x) - 5 = 0
⇒ 2x + 2/x - 5 = 0
⇒ 2x² - 5x + 2 = 0
Splitting the middle term
⇒ 2x² - 4x - x + 2 = 0
⇒ 2x(x - 2) - 1(x - 2) = 0
⇒ (2x - 1)(x - 2) = 0
⇒ 2x - 1 = 0 or x - 2 = 0
⇒ 2x = 1 or x = 2
⇒ x = 1/2 or x = 2
Therefore the number is 2 or 1/2.