# The value of isA. 4B. 3C. – 2D. 3.5

In given equation let x =

So, x = √(6 + x)

Now squaring both side

x2 = 6 + x

X2 – x – 6 = 0

X2 – 3x + 2x – 6 = 0

x(x – 3) + 2(x – 3) = 0

(x – 3) (x + 2) = 0

x = 3 or – 2

x cannot be equal to – 2 as root can never be negative.

x = 3

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