Advertisements
Advertisements
Question
Solve the following quadratic equations by factorization:
`x^2+(a+1/a)x+1=0`
Solution
We have been given
`x^2+(a+1/a)x+1=0`
`x^2+ax+1/ax+1=0`
`x(x+a)+1/a(x+a)=0`
`(x+1/a)(x+a)=0`
Therefore,
`x+1/a=0`
`x=-1/a`
or,
x + a = 0
x = -a
Hence, `x=-1/a` or x = -a
APPEARS IN
RELATED QUESTIONS
Solve the following quadratic equations by factorization:
6x2 - x - 2 = 0
Solve the following quadratic equations by factorization:
`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`, x ≠ 2, 4
Solve:
`1/p + 1/q + 1/x = 1/(x + p + q)`
Solve the following quadratic equation by factorisation.
x2 + x – 20 = 0
The number of quadratic equations having real roots and which do not change by squaring their roots is
Solve the following equation :
`1/(("x" - 1)(x - 2)) + 1/(("x" - 2)("x" - 3)) + 1/(("x" - 3)("x" -4)) = 1/6`
Solve the following equation by factorization
`(2)/(3)x^2 - (1)/(3)x` = 1
Solve the following equation by factorization
`(1)/(7)(3x – 5)^2`= 28
Find the values of x if p + 7 = 0, q – 12 = 0 and x2 + px + q = 0,
There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?
