Advertisements
Advertisements
प्रश्न
Solve the following equation by factorization
`(1)/(2a + b + 2x) = (1)/(2a) + (1)/b + (1)/(2x)`
Advertisements
उत्तर
`(1)/(2a + b + 2x) = (1)/(2a) + (1)/b + (1)/(2x)`
⇒ `(1)/(2a + b + 2x) = (1)/(2x) + (1)/(2a) + (1)/(b)`
⇒ `(2x - (2a + b + 2x))/((2a + b + 2x)2x) = (b + 2a)/(2ab)`
⇒ `(-(2a + b))/((2a + b + 2x)2x) = ((2a + b))/(2ab)`
⇒ `(-1)/((2a + b + 2x)2x) = (1)/(2ab)`
⇒ -2ab = (2a + b + 2x)2x
⇒ 4ax + 2xb + 4x2 = -2ab
⇒ 4x2 + 2bx + 4ax + 2ab = 0
⇒ 2x(2x + b) + 2a(2x + b) = 0
⇒ (2x + 2a)(2x + b) = 0
⇒ 2x + 2a = 0 or 2x + b = 0
`x = -a or x = -b/(2)`
Hence, the roots of the given equation are
`-a and -b/(2)`.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
`1/(x+4)-1/(x-7)=11/30` , x ≠ 4, 7
There are three consecutive integers such that the square of the first increased by the product of the first increased by the product of the others the two gives 154. What are the integers?
Solve the following quadratic equation by factorisation.
\[6x - \frac{2}{x} = 1\]
If \[\left( a^2 + b^2 \right) x^2 + 2\left( ab + bd \right)x + c^2 + d^2 = 0\] has no real roots, then
Solve the following equation: `("x" + 3)/("x" + 2) = (3"x" - 7)/(2"x" - 3)`
A two digit number is 4 times the sum of its digit and twice the product of its digit. Find the number.
In each of the following determine whether the given values are solutions of the equation or not
2x2 - 6x + 3 = 0; x = `(1)/(2)`
Solve the following equation by factorization
5x2 – 8x – 4 = 0 when x∈Q
The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form ail equation in x and solve it to find the numbers.
Using quadratic formula find the value of x.
p2x2 + (p2 – q2)x – q2 = 0
