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√(x² -4x +3) + √(x²-9) = √(4x²-14x+16)
question ask 3 is a root of equation ,
first we find condition ,
x²-4x +3 ≥0 => x≥3 and x≤1
x²-9 ≥0 => x≥3 and x≤ -3
4x²-14x+16≥0 => 2x²-7x+8≥0
Discriminant = 49 -64<0
so , x €R
take common interval here ,
x≥3 and x≤ -3
now,
√(x²-4x+3) + √(x²-9) = √(4x²-14x+16)
here 3 is a root of equation ,
put x = 3
LHS = √(9-12+3) +√9-9) =0
RHS = √(36-42+16) =√10
LHS ≠ RHS
3 is not a root of given equation ,
question ask 3 is a root of equation ,
first we find condition ,
x²-4x +3 ≥0 => x≥3 and x≤1
x²-9 ≥0 => x≥3 and x≤ -3
4x²-14x+16≥0 => 2x²-7x+8≥0
Discriminant = 49 -64<0
so , x €R
take common interval here ,
x≥3 and x≤ -3
now,
√(x²-4x+3) + √(x²-9) = √(4x²-14x+16)
here 3 is a root of equation ,
put x = 3
LHS = √(9-12+3) +√9-9) =0
RHS = √(36-42+16) =√10
LHS ≠ RHS
3 is not a root of given equation ,
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