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प्रश्न
`(x/(x+1))^2-5(x/(x+1)+6=0,x≠b,a`
उत्तर
`(x/(x+1))^2-5(x/(x+1))+6=0`
Putting `x/(x+1)=y`, We get:
⇒`y^2-5y+6=0`
⇒`y^2-5y+6=0`
⇒`y^2-(3+2)y+6=0`
⇒`y^2-3y-2y+6=0`
⇒`y(y-3)-2(y-3)=0`
⇒`(y-3)(y-2)=0`
⇒`y-3=0 or y-2=0`
`y=3 or y=2` Case I:
If y = 3, we get
⇒`x/(x+1)=3`
⇒x=3(x+1) [On cross multiplying
⇒`x=3x+3`
⇒`x=(-3)/2`
Case II:
If y =2, we get:
⇒`x/(x+1)=2`
⇒`x=2(x+1) `
⇒`x=2x+2`
⇒`-x=2`
⇒`x=-2`
Hence, the roots of the equation are `-3/2` and -2
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