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The equation (x^2+x=1)^2+1=(x^2+x+1)(x^2...

The equation `(x^2+x=1)^2+1=(x^2+x+1)(x^2-x-5)` for `x in (-2,3)` will have number of solutions. `1` b. `2` c. `3` d. 0

A

1

B

2

C

3

D

zero

Text Solution

Verified by Experts

The correct Answer is:
4
`(x^(2) + x + 1)^(2) + 1 = (x^(2) + x + 1) (x^(2) - x - 5)`
Since `x^(2) + x + 1 gt 0 AA x in R,`
Dividing both sides by `x^(2) + x + 1`, we get
`(x^(2) + x + 1) + (1)/((x^(2) + x + 1)) = x^(2) - x - 5`
`therefore L.H.S. ge 2`
But `x^(2) - x - 5 lt 1` for `x in (-2, 3)`
Therefore, the equation has no solution.
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