Advertisements
Advertisements
प्रश्न
The sum of a number and its reciprocal is `2 9/40`. Find the number.
Advertisements
उत्तर
Let the number be A. Then as per the question,
`"A" + 1/ "A" = 2 9/40`
⇒ `"A"^2 + 1 - 89/40 "A" = 0`
⇒ 40 A2 + 40 - 89 A=0
⇒ 40A2 - 89A + 40 = 0
⇒ 40 A2 - 64A - 25A + 40 = 0
⇒ 8A (5A - 8) - 5 (5A-8) = 0
⇒ (8A - 5)(5A - 8) = 0
⇒ A = `5/8 , 8/5`
APPEARS IN
संबंधित प्रश्न
Solve for x: `(x-3)/(x-4)+(x-5)/(x-6)=10/3; x!=4,6`
Solve the following quadratic equations by factorization:
`(2x)/(x-4)+(2x-5)/(x-3)=25/3`
A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.
Solve 2x2 – 9x + 10 =0; when x ∈ Q
Solve:
x(x + 1) + (x + 2)(x + 3) = 42
Find the values of k for which the roots are real and equal in each of the following equation:
\[kx\left( x - 2\sqrt{5} \right) + 10 = 0\]
Three consecutive natural numbers are such that the square of the first increased by the product of other two gives 154. Find the numbers.
The hypotenuse of a right-angled triangle is 17cm. If the smaller side is multiplied by 5 and the larger side is doubled, the new hypotenuse will be 50 cm. Find the length of each side of the triangle.
Solve equation using factorisation method:
(x + 1)(2x + 8) = (x + 7)(x + 3)
The product of two successive integral multiples of 5 is 300. Then the numbers are:
