Advertisements
Advertisements
प्रश्न
Solve the following equation by factorization
`x/(x + 1) + (x + 1)/x = (34)/(15)`
Advertisements
उत्तर
`x/(x + 1) + (x + 1)/x = (34)/(15)`
`(x^2 + x^2 + 2x + 1)/(x(x + 1)) = (34)/(15)`
⇒ `(2x^2 + 2x + 1)/(x^2 + x) = (34)/(15)`
⇒ 30x2 + 30x + 15 = 34x2 + 34x
⇒ 30x2 + 30x + 15 - 34x2 - 34x = 0
⇒ -4x2 - 4x + 15 = 0
⇒ 4x2 + 4x - 15 = 0
⇒ 4x2 + 10x - 6x - 15 = 0
⇒ 2x(2x + 5) - 3(2x + 5) = 0
⇒ (2x + 5) (2x - 3) = 0
Either 2x + 5 = 0,
then 2x = -5
⇒ x = `(-5)/(2)`
or
2x - 3 = 0,
then 2x = 3
⇒ x = `(3)/(2)`
Hence x = `(-5)/(2), (3)/(2)`.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation by
factorisation.
5m2 = 22m + 15
If 1 is a root of the quadratic equation \[3 x^2 + ax - 2 = 0\] and the quadratic equation \[a( x^2 + 6x) - b = 0\] has equal roots, find the value of b.
If the equation x2 + 4x + k = 0 has real and distinct roots, then
A two digit number is four times the sum and 3 times the product of its digits, find the number.
Find two natural numbers which differ by 3 and whose squares have the sum of 117.
In each of the following, determine whether the given values are solution of the given equation or not:
`x = 1/x = (13)/(6), x = (5)/(6), x = (4)/(3)`
Solve the following equation by factorization
(x – 3) (2x + 5) = 0
Solve the following equation by factorization
3x2 – 5x – 12 = 0
Solve the following equation by factorization
`x + (1)/x = 2(1)/(20)`
The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.
