Number of solutions of the equation $(x^{2} + x + 1)^{2} - 4 = (x^{2} + x + 1) (x^{2} - x - 5)$ for $x \in (- 2, 3)$ , are:

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2$
${(x^{2} + x + 1)}^{2} - 4 = (x^{2} + x + 1) (x^{2} - x - 5) \Rightarrow x^{4} + x^{2} + 1 + 2 x^{3} + 2 x^{2} + 2 x - 4 = x^{4} - x^{3} - 5 x^{2} + x^{3} - x^{2} - 5 x + x^{2} - x - 5 \Rightarrow 2 x^{3} + 8 x^{2} + 8 x + 2 = 0$
$\Rightarrow x^{3} + 4 x^{2} + 4 x + 1 = 0$
$\Rightarrow (x + 1) (x^{2} + 3 x + 1) = 0$
As $x^{2} + 3 x + 1$ has one solution in the range.
Also, $x + 1 = 0 \Rightarrow x = - 1$ .
Hence, number of solutions between $(- 2, 3)$ is: $2$ .

Suggest Corrections

Similar questions

ϵ

(x^{2} + x + 1)^{2} + 1 = (x^{2} + x + 1) (x^{2} + x + 5)

x^{2} - 3 x + 2 = 0, x = 2, x = - 1

x^{2} + x + 1 = 0, x = 0, x = 1

x^{2} - 3 \sqrt{3} x + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}

x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{4}{3}

2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3

x^{2} - \sqrt{2} x - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2}

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}

x^{2} - 5 | x | + 6 = 0

x^{2} - 7 x + 12 = 0 : x = 3, x = 4

{log}_{\sqrt{3} + ∣ ∣ x^{2} - 1 ∣ ∣} (3 - x^{2} - \frac{1}{x^{2}}) = {log}_{\sqrt{3} + ∣ ∣ x^{2} - 1 ∣ ∣} (5 + x^{2} - x)

Join BYJU'S Learning Program