Advertisements
Advertisements
प्रश्न
Solve the following equation by factorization
`(1)/(x - 3) - (1)/(x + 5) = (1)/(6)`
उत्तर
`(1)/(x - 3) - (1)/(x + 5) = (1)/(6)`
`(x + 5 - x + 3)/((x - 3) (x + 5)) = (1)/(6)`
⇒ `(8)/(x^2 + 2x - 15) = (1)/(6)`
⇒ x2 + 2x - 15 = 48
⇒ x2 + 2x - 15 - 48 = 0
⇒ x2 + 2x - 63 = 0
⇒ x2 + 9x - 7x - 63 = 0
⇒ x(x + 9) - 7(x + 9) = 0
⇒ (x + 9) (x - 7) = 0
Either x + 9 = 0,
then x = -9
or
x - 7 = 0,
then x = 7
Hence x = -9, 7.
APPEARS IN
संबंधित प्रश्न
Solve each of the following equations by factorization:
`9/2x=5+x^2`
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 one root the equation 2x2 + kx + 4 = 0 is 2, then the other root is
A two digit number is such that the product of the digit is 12. When 36 is added to the number, the digits interchange their places. Find the numbers.
Solve equation using factorisation method:
(x + 1)(2x + 8) = (x + 7)(x + 3)
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.
In each of the following, determine whether the given values are solution of the given equation or not:
x2 - 3`sqrt(3)` + 6 = 0; x = `sqrt(3)`, x = -2`sqrt(3)`
Solve the following equation by factorization
x(6x – 1) = 35
Solve the following equation by factorization
x2– 4x – 12 = 0,when x∈N
If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is `13/21` more than the product of the roots, then find the value of k.