Advertisements
Advertisements
प्रश्न
Solve the following by reducing them to quadratic equations:
`sqrt(x/(1 -x)) + sqrt((1 - x)/x) = (13)/(6)`.
Advertisements
उत्तर
Given equation `sqrt(x/(1 -x)) + sqrt((1 - x)/x) = (13)/(6)`
Putting `sqrt(x/(1 - x)) = y,` then given equation reducible to the form `y + (1)/y = (13)/(6)`
⇒ `(y^2 + 1)/y = (13)/(6)`
⇒ 6y2 + 6 = 13y
⇒ 6y2 - 13y + 6 = 0
⇒ 6y2 - 9y - 4y + 6 = 0
⇒ 3y(2y - 3) -2(2y - 3) = 0
⇒ (2y - 3) (3y - 2) = 0
⇒ 2y - 3 = 0 or 3y - 2 = 0
⇒ y = 3/2 or y = 2/3
But `sqrt(x/(1 - x)) = y`
∴ `sqrt(x/(1 - x)) = (3)/(2)`
Squaring `x/(1 -x) = (9)/(4)`
⇒ 4x = 9 - 9x
⇒ 13x = 9
⇒ x = `(9)/(13)`
or
`sqrt(x/(1 - x)) = (2)/(3)`
Squaring `x/(1 - x) = (4)/(9)`
⇒ 9x = 4 - 4x
⇒ 9x + 4x = 4
⇒ 13x = 4
⇒ x = `(4)/(13)`
Hence, the required roots are `{9/13,4/13}`.
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`, x ≠ 2, 4
The sum of a number and its reciprocal is 17/4. Find the number.
A two digit number is such that the product of the digits is 16. When 54 is subtracted from the number the digits are interchanged. Find the number
Find the value of k for which the following equations have real and equal roots:
\[\left( k + 1 \right) x^2 - 2\left( k - 1 \right)x + 1 = 0\]
Two natural numbers differ by 4. If the sum of their square is 656, find the numbers.
Divide 29 into two parts so that the sum of the square of the parts is 425.
Solve the following equation by factorization
2x2 – 9x + 10 = 0,when x∈N
Solve the following equation by factorization
`(x + 2)/(x + 3) = (2x - 3)/(3x - 7)`
The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by `(4)/(35)`. Find the fraction.
Solve the following equation by factorisation :
x2 + 6x – 16 = 0
