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2√2xˆ2 + 2√5x = 0
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\[2\sqrt{2^{2}}+2\sqrt{5x}\] we have some radicals, but the proceidure requires factoring, Want me to guido you through it?
the problem is \[2\sqrt{2x ^{2}}+2\sqrt{5x}=0\] and yes please
much better: Let's begin by applying a common factor: \[2(\sqrt{2x ^{2}}+\sqrt{5x})=0\]
we would pull out an x
\[\sqrt{x}*(2\sqrt{2}*\sqrt{x}+2\sqrt{5})=0\]
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x^1/2=0 OR \[2\sqrt{2x}=-2\sqrt{5}\]
on squareing both sides \[4*2x=4*5\]
8x=20 x=5/2
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