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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 equal to-A. 51B. -51C. -126D. 126 |
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Answer» Correct Answer - C r is root, therefore `r^(2)+2r+6=0` `(r+2)(r+3)(r+4)(r+5)=(r^(2)+2r+6+3r)(r^(2)+4r+5r+20)` `=(3r)(7r+14)` `21(r^(2)+2r)=21(-6)=-126` |
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