Question

14. Find the side of a cube whose volume is x3 + 6x2 + 12x + 8 cubic units.
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Answer 1

Step-by-step explanation:

Let the side of the cube be x.

W.K.T volume of a cube = x^3

Given that, Volume =

${x}^{3} + 6 {x}^{2} + 12x + 8$

Therefore, by substituting x in the volume formula,

${x}^{3} = {x}^{3} + 6 {x}^{2} + 12x + 8$

$= > 6 {x}^{2} + 12x + 8 = 0$

Comparing the equation with general form, we get a = 6, b = 12, c = 8,

Substituting a,b,c in the quadratic formula, we get,

$\frac{ - 12 + - \sqrt{ {12}^{2} - 4(6)(12)(8) } }{2(6)}$

$\frac{ - 12 + - \sqrt{144 - 2304} }{12}$

$\frac{ - 12 + - \sqrt{2160} }{12}$

$\frac{ - 12 + - 12 \sqrt{15} }{12}$

$- 1 + \sqrt{15} \: \: and \: - 1 - \sqrt{15}$

Length of side of a cube cannot be negative... Hence we discard the value

$- 1 - \sqrt{15}$

Therefore length of side of cube is

$- 1 + \sqrt{15} \: \: \: units$

Equating further we get,

$- 1 + 3.872$

$= > 2.872 \: \: units$

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