Question

if m=3-√5; prove m²-5m+√5+1=0

Answer 1

Given that :

$m = 3 - \sqrt{5}$

No, we have to prove that :

${m}^{2} - 5m + \sqrt{5} + 1 = 0$

On taking LHS :

${(3 - \sqrt{5}) }^{2} - 5(3 - \sqrt{5} ) + \sqrt{5} + 1 \\ \\ = > 9 - 6 \sqrt{5} + 5 - 15 + 5 \sqrt{5} + \sqrt{5} + 1 \\ \\ = > 15 - 15 - 6 \sqrt{5} + 5 \sqrt{5} + \sqrt{5} \\ \\ = > - 6 \sqrt{5} + 6 \sqrt{5} \\ \\ = > 0 = RHS \\ \\ HENCE \: PROVED$

Answer 2

Answer:

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Step-by-step explanation:

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