Question

(16-x ²) (x²+4) (x²+x+1)(x²-x-3) < = 0 so solve he inequality​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

We have,

$(16 - {x}^{2} )( {x}^{2} + 4)( {x}^{2} + x + 1)( {x}^{2} - x - 3) \leqslant 0$

Now, (x²+4) is always positive. Also, x²+x+1 is positive.

so,

(16 - x²)(x² - x - 3) must be less than 0

$(16 - {x}^{2} )( {x}^{2} - x - 3) \leqslant 0$

$= > ( {x}^{2} - 16)( {x}^{2} - x - 3) \geqslant 0$

$= > (x + 4)(x - 4)(x - \frac{1 + \sqrt{13} }{2} )(x - \frac{1 - \sqrt{13} }{2} ) \geqslant 0$

$= > x \: \: belongs \: \: to \: ( - \infty \: \: to \: \: - 4] \: \: union \: \: [ \frac{1 - \sqrt{13} }{2} \: \: to \: \: \frac{1 + \sqrt{13} }{2} ] \: \: union \: \: [4 \: \: to \: \: \infty )$