What is the value of the positive square root | vedclass

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Question

(A) To find the square root of $(69 + 28 \sqrt{5})$,we assume it is of the form $(a + b \sqrt{5})$.Squaring both sides,we get: $(a + b \sqrt{5})^2 = 6

Vedclass · Accepted Answer

$7 + 2 \sqrt{5}$

Vedclass · Answer

(A) To find the square root of $(69 + 28 \sqrt{5})$,we assume it is of the form $(a + b \sqrt{5})$.Squaring both sides,we get: $(a + b \sqrt{5})^2 = 69 + 28 \sqrt{5}$.Expanding the left side: $a^2 + 5b^2 + 2ab \sqrt{5} = 69 + 28 \sqrt{5}$.Comparing the rational and irrational parts:$a^2 + 5b^2 = 69$ and $2ab = 28$,which means $ab = 14$.Possible pairs for $(a, b)$ such that $ab = 14$ are $(14, 1), (7, 2), (2, 7), (1, 14)$.Testing $(a, b) = (7, 2)$:$a^2 + 5b^2 = 7^2 + 5(2^2) = 49 + 5(4) = 49 + 20 = 69$.This satisfies the equation.Thus,the square root is $(7 + 2 \sqrt{5})$.

What is the value of the positive square root of $(69 + 28 \sqrt{5})$?

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