First slide

Theory of equations

Question

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,

Moderate

A

1
B

2
C

3
D

0

Solution

${(x^{2} + x + 1)}^{2} + 1 = (x^{2} + x + 1) (x^{2} - x - 5)$
Since, $x^{2} + x + 1 > 0 \forall x \in R$
Dividing both sides by $x^{2} + x + 1$ , we get
$(x^{2} + x + 1) + \frac{1}{(x^{2} + x + 1)} = x^{2} - x - 5$
$∴ L.H.S. \geq 2$
But $x^{2} - x - 5 < 1 for x \in (- 2, 3)$
Therefore, the equation has no solution.

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