Find the root of 3x2-4root3x+4=0 by completing square method
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3x²-4√3x+4=0
Dividing LHS by 3,we get
x²-[(4√3)/3]+4/3=0
=>x²-(4/√3)+4/3=0
=>x²-4/√3 = 4/3
Adding (1/2 of coefficient of x)² on both sides,we get,
x²-(4/√3)+(1/2*4/√3)²=(4/3)+(1/2*4/√3)²
=>x²-(4/√3)+(4/2√3)²=(4/3)+(4/2√3)²
=>{x-(4/2√3)}²=(4/3)+(16/12)=(4/3)+(4/3)=8/3
=>x-(4/2√3)=√(8/3)
=>x=2(4/2√3).
Dividing LHS by 3,we get
x²-[(4√3)/3]+4/3=0
=>x²-(4/√3)+4/3=0
=>x²-4/√3 = 4/3
Adding (1/2 of coefficient of x)² on both sides,we get,
x²-(4/√3)+(1/2*4/√3)²=(4/3)+(1/2*4/√3)²
=>x²-(4/√3)+(4/2√3)²=(4/3)+(4/2√3)²
=>{x-(4/2√3)}²=(4/3)+(16/12)=(4/3)+(4/3)=8/3
=>x-(4/2√3)=√(8/3)
=>x=2(4/2√3).
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