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X^4+14x^2-32=0 Find all the imaginary solutions by factoring
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\[x^4+14x^2-32=0\]let x^2=a
\[a^2+14a-32=0\]\[(a + 14)(a - 2 )=0\]\[a=-14~~~~~~and~~~~~~a=2\]
\[x^2=a~~~~~~~~so\]\[x^2=-14~~~~~~~x=±i \sqrt{14}\]\[x^2=2~~~~~~~x=± \sqrt{2}\]
Why would you let x^2=a
\[x ^{4}+14x^2-32=0\]\[(x^2+16)(x^2-2)\] \[x^2=-16\]\[\sqrt{x}=\sqrt{-16}\]\[x=\pm4i\] \[x^2=2\]\[\sqrt{x}=\sqrt{2}\]\[x=\pm \sqrt{2}\]
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Oh thank you That makes more sense
yes true it does, it become alg 1 from college alg. by simply saying let x^2=a
no problem
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