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If y varies directly as the square root of (x + 1) and y = 6 when x = 3. Find x when y = 9.

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y ∝ √(x + 1)

y = k√(x + 1) --------- eqn(1)

where k is the proportionality constant.

divide both sides by √(x + 1)

k = y/√(x + 1)

if y = 6 and x = 3,

k = 6/√(3 + 1)

k = 6/√4

k = 6/2

k = 3.

Similarly, if y = 9, k = 3 (because it is constant for any values of x and y)

from eqn(1), 9 = 3√(x + 1)

divide both sides by 3

9/3 = √(x + 1)

3 = √(x + 1)

take the square of both sides to cancel out the square root on the right hand side.

3^{2} = x + 1

9 = x + 1

subtract 1 from both sides,

9 - 1 = x

8 = x

Thus, x = 8

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