Question

a=√5-1,then( a^4+4a^3+6a^2+4a-6)=?
$a = \sqrt{5 - 1then(a {4 \time}^{2}$
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Answer 1

Answer:

supposed to be the right answer in one method,

If a-1/a=5 the a-1=5a

-1=5a-a which is -1=4a which is a=-1/4 ————1

a²-1/a² is (-1/4)² -1/(-1/4)² which is 1/16 -1/1/16.

-15/16 / 1/16

Applying another method from given condition in problem.

Other right answer is a-1/a=5—Eq1

So squaring both sides

We get a² -2a/a +1/a² which is a² +1/a² -2=5×5

Which further is a²+1/a²-2=25—Eq2

Now adding 4 to both sides

a²+1/a²-2+4=25+4

Which is a²+2+1/a²=29

Further square root of both sides

√a²+2+1/a²=√29

a+1/a=√29

Now in problem the then asked what is a²-1/a² is (a+1/a)(a-1/a)

a+1/a ×a-1/a

Whereas in question a-1/a=5 and a+1/a=√29 is proved

So +-5√29 is Answer.