Question

please solve this question as soon as possible!

Answer 1

Let A be the arthmetic mean and G be the geometric mean between a and b
so...A=2G
therefore
${a + b} \div 2 = 2 \sqrt{ab}$
${a + b} = 4 \sqrt{ab}$
${a + b} \div \sqrt{ab } = 4$
we can write
$a \div \sqrt{ab} + b \div \sqrt{ab } = 4$

$\sqrt{a \div b} + \sqrt{b \div a} = 4$
let a/b =k
$\sqrt{k} + \sqrt{1 \div k} = 4$
squaring both side we get
k+1/k+2=16
multiply by k both side be get
k^2+14k+1=0
solve for k we get
$k = 7 + 4 \sqrt{3}$
or
$k = 7 - 4 \sqrt{3}$
put the value of k a/b
HENCE PROVED


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