Question

if a=2-√5/2+√5and b =2+√5/2-√5 find a^2+b^2​

Answer 1

Step-by-step explanation:

$a=\frac{2-\sqrt{5} }{2+\sqrt{5} }$

$b=\frac{2+\sqrt{5} }{2-\sqrt{5} }$

Now we have to rationalize the denominator of both a and b.

$a=\frac{2-\sqrt{5} }{2+\sqrt{5} } *\frac{2-\sqrt{5} }{2-\sqrt{5} } =\frac{4-2\sqrt{5} -2\sqrt{5}+5 }{2^{2}-\sqrt{5} ^{2} } =\frac{9-4\sqrt{5} }{-1} =-(9-4\sqrt{5} )=-9+4\sqrt{5}$

$b=\frac{2+\sqrt{5} }{2-\sqrt{5} } *\frac{2+\sqrt{5} }{2+\sqrt{5} } =\frac{4+2\sqrt{5}+2\sqrt{5}+5 }{2^{2}-\sqrt{5} ^{2} } =\frac{9+4\sqrt{5} }{-1} =-(9+4\sqrt{5} )=-9-4\sqrt{5}$

$a^{2} +b^{2} =(-9+4\sqrt{5} )^{2} +(-9-4\sqrt{5} )^{2} \\=[(-9)^{2} +2(-9)(4\sqrt{5} )+(4\sqrt{5} )^{2} ]+[(-9)^{2} -2(-9)(4\sqrt{5} )+(4\sqrt{5} )^{2} ]\\=(81-72\sqrt{5} +80)+(81+72\sqrt{5} +80)\\=(161-72\sqrt{5} )+(161+72\sqrt{5} )\\=161-72\sqrt{5} +161+72\sqrt{5} \\=161+161-72\sqrt{5} +72\sqrt{5} \\=322$

Answer = 322

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