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In how many points the line `y + 14 = 0` cuts the curve whose equation is `x(x^(2) + x + 1) + y = 0 ? ` |
Answer» Correct Answer - one When graphs of `y - 14 and y = -x(x^(2) + x + 1) = 0` intersect, we have `- (x^(3) + x^(2) + x ) = - 14 or x^(3) + x^(3) + x - 14 = 0` Now x = 2 satisfies the equation,then one root is x = 2. Dividing `x^(2) + x^(2) + x + 7) = 0` we have `(x - 2) (x^(2)+ 3x + 7 ) = 0` `rArr x = 2 or x^(2) + 3x + 7 = 0` Now `x ^(2) + 3x + 7 = 0` has non-real roots, Hence, graphs cut in only one real point. |
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