find the roots of the equation X+√(X-2) =4
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x + √(x - 2) = 4
√(x - 2) = 4 -x
take square Both sides
x - 2 = 16 -8x + x²
x² -9x +18 = 0
x² -6x -3x +18 =0
x(x -6) -3(x -6) =0
(x -6)(x -3) =0
x = 6 , 3
according to √(x - 2) and (4 -x) ≥0
x≥2 and x ≤4
so, x ≠ 6
only x = 3
√(x - 2) = 4 -x
take square Both sides
x - 2 = 16 -8x + x²
x² -9x +18 = 0
x² -6x -3x +18 =0
x(x -6) -3(x -6) =0
(x -6)(x -3) =0
x = 6 , 3
according to √(x - 2) and (4 -x) ≥0
x≥2 and x ≤4
so, x ≠ 6
only x = 3
234Nishant1:
ty
bro x=6 is extentioneous root....it is not the root of this equation put x=6 in real equation u will find that 6 is not root of the eqn...only x= 3 satisfy so only x= 3 is root of equation
ohh
ty
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