anonymous
  • anonymous
Please help me factor this biquadratic expression: a^4 - 3a^2 + 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Directrix
  • Directrix
a^2 - a - 1) ( a^2 +a -1)
ash2326
  • ash2326
We've got \[a^4-3a^2+1\] let's substitute a^2=u so a^4=u^2 now we have \[u^2-3u+1\] now we'll use quadratic formula to figure out the factors for \[ax^2+bx+c\] the roots are given as \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] here we have u^2-3u+1=0 a=1, b=-3 and c=1 so \[u=\frac{3\pm\sqrt{3^2-4*1*1}}{2*1}\] or \[u=\frac{3\pm\sqrt{5}}{2}\] so we'll have the factors as \[(u-(\frac{3+\sqrt{5}}{2}))(u-(\frac{3-\sqrt{5}}{2}))\] now we have u=a^2 , so the factors are \[(a^2-(\frac{3+\sqrt{5}}{2}))(a^2-(\frac{3-\sqrt{5}}{2}))\]
anonymous
  • anonymous
\[ a^4 - 3a^2 + 1 = (a^2-1)^2 -a^2 =\left(a^2-1+a\right) \left(a^2-1-a\right) \]

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anonymous
  • anonymous
Got it. Thanks a lot.
Directrix
  • Directrix
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