Advertisements
Advertisements
प्रश्न
`1/x-1-1/(x+5)=6/7,x≠1,-5`
उत्तर
`1/x-1-1/(x+5)=6/7,x≠1,-5`
⇒` (x+5-x+1)/((x-1) (x+5))=6/7`
⇒`6/(x^2+4x-5)=6/7`
⇒`x^2+4x-5=7`
⇒`x^2+4x-12=0`
⇒`x^2+6x-2x-12=0`
⇒`x(x+6)-2(x+6)=0`
⇒`(x+6)(x-2)=0`
⇒`x+6=0 or x-2=0`
⇒` x=-6 or x=2`
Hence, -6 and 2 are the roots of the given equation.
APPEARS IN
संबंधित प्रश्न
(x + 5)(x - 2) = 0, find the roots of this quadratic equation
Check whether the following is quadratic equation or not.
(x + 2)3 = x3 − 4
In the following, determine whether the given values are solutions of the given equation or not:
`x^2 - 3sqrt3x+6=0`, `x=sqrt3`, `x=-2sqrt3`
Without solving, comment upon the nature of roots of the following equation:
x2 – ax – b2 = 0
`sqrt3x^2+11x+6sqrt3=0`
`a^2b^2x^2+b^2x-1=0`
`1/(x+1)+2/(x+2)=5/(x+4),x≠-1,-2,-4`
Write the following equation in the form ax2 + bx + c = 0, then write the values of a, b, c for the equation.
2y = 10 – y2
`2/3`and 1 are the solutions of equation mx2 + nx + 6 = 0. Find the values of m and n.
Solve the following equation by using formula :
25x2 + 30x + 7 = 0
