Question

$p(x) = {x}^{4} - 16$
please help me...

Answer 1

srry ....it's not x= +-2
it's x= +√4,-√4
hope this helps you....

Answer 2

Hello Friend.....

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The answer of u r question is.....

Ans:

Given,

$p(x) = {x}^{4} - 16$

$= {x}^{4} - 16 = 0$

$= ( {x}^{2 {)}^{2} } - {4}^{2} = 0$

$= ( {x}^{2} + 4)( {x}^{2} - 4) = 0$
$( {x}^{2} + 4) = 0 \: \: or \: ( {x}^{2} - 4) = 0$

$= {x}^{2} - 4 \: or \: {x}^{2} = 4$

$x = \sqrt{4}$

$x + \sqrt{ - 4} x = + 2$
x=2,2-x,x=√-4,√-4...


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