If the volume of cuboid is 2x^2-16,find the dimensions
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hello friend
Let V is equal to 0 i.e
![2 {x}^{2} - 16 = 0 \\ { (\sqrt{2x}) }^{2} - {4}^{2} = 0 \\ ( \sqrt{2x} + 4)( \sqrt{2x} - 4) = 0 \\ ( \sqrt{2x} + 4)(( { \sqrt[4]{2x}) }^{2} - {2}^{2} ) = 0 \\ ( \sqrt{2x } + 4)( \sqrt[4]{2x} - 2)( \sqrt[4]{2x} + 2) = 0](https://tex.z-dn.net/?f=2+%7Bx%7D%5E%7B2%7D+++-+16+%3D+0+%5C%5C+++%7B+%28%5Csqrt%7B2x%7D%29++%7D%5E%7B2%7D++-++%7B4%7D%5E%7B2%7D++%3D+0+%5C%5C+%28+%5Csqrt%7B2x%7D++%2B+4%29%28+%5Csqrt%7B2x%7D++-+4%29+%3D+0+%5C%5C+%28+%5Csqrt%7B2x%7D+%2B+4%29%28%28+%7B+%5Csqrt%5B4%5D%7B2x%7D%29+%7D%5E%7B2%7D+++-++%7B2%7D%5E%7B2%7D+%29+%3D+0+%5C%5C+%28+%5Csqrt%7B2x+%7D++%2B+4%29%28+%5Csqrt%5B4%5D%7B2x%7D++-+2%29%28+%5Csqrt%5B4%5D%7B2x%7D+%2B+2%29+%3D+0+ "2 {x}^{2} - 16 = 0 \\ { (\sqrt{2x}) }^{2} - {4}^{2} = 0 \\ ( \sqrt{2x} + 4)( \sqrt{2x} - 4) = 0 \\ ( \sqrt{2x} + 4)(( { \sqrt[4]{2x}) }^{2} - {2}^{2} ) = 0 \\ ( \sqrt{2x } + 4)( \sqrt[4]{2x} - 2)( \sqrt[4]{2x} + 2) = 0")
those 3 factors represent dimensions..
Hope it helps ..
Let V is equal to 0 i.e
those 3 factors represent dimensions..
Hope it helps ..
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