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Let r be a root of the equation x^2+2x+6...

Let `r` be a root of the equation `x^2+2x+6=0`. The value of `(r+2)(r+3)(r+4)(r+5)` is

Text Solution

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The given eqation is,
`x^2+2x+6=0`
If r is one of the root of this equation,then `r^2+2r+6=0`
`(r^2+2r)=-6 and r^2=-2r-6`
Now, `(r+2)(r+3)(r+4)(r+5)=(r^2+5r+6)(r^2+9r+20)`
`=(-2r-6+5r+6)(-2r-6+9r+20)`
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