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In how many points the line y + 14 = 0 c...

In how many points the line `y + 14 = 0` cuts the curve whose equation is `x(x^(2) + x + 1) + y = 0 ? `

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Verified by Experts

The correct Answer is:
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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