solve by thecompleting of square method root2x^2-3x-2root2
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2x^2-3x- 2 root 2 = 0
(1st step -- divide the whole equation by coefficient of x^2 =2)
x^2-3/2x - root 2 = 0
(2nd step-- transfer the constant term)
x^2-3/2x=root 2
☆now the most complicated step comes
(3rd step-- by adding both side ( 1/2 of coefficient of x)^2) mean
(3/2×1/2)^2=> (3/4)^2
x^2-3/2x+[3/4]^2=root 2+[3/4]^2
now there is looking like an identity of (a-b)^2=a^2+b^2-2ab
(x-3/4)^2=root 2 (3/4)^2
now ,
(x-3/4)^2=root 2+9/16
(now 4th step --taking square root both sides )
x-3/4=plus minus in root root 2+9/16
x=in root root2+9/16+3/4
aage please aap solve kar dijiyega............
2x^2-3x- 2 root 2 = 0
(1st step -- divide the whole equation by coefficient of x^2 =2)
x^2-3/2x - root 2 = 0
(2nd step-- transfer the constant term)
x^2-3/2x=root 2
☆now the most complicated step comes
(3rd step-- by adding both side ( 1/2 of coefficient of x)^2) mean
(3/2×1/2)^2=> (3/4)^2
x^2-3/2x+[3/4]^2=root 2+[3/4]^2
now there is looking like an identity of (a-b)^2=a^2+b^2-2ab
(x-3/4)^2=root 2 (3/4)^2
now ,
(x-3/4)^2=root 2+9/16
(now 4th step --taking square root both sides )
x-3/4=plus minus in root root 2+9/16
x=in root root2+9/16+3/4
aage please aap solve kar dijiyega............
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