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solve 5x^2-7x+2=0 completing the square. what is the constant added on to form the perfect square trinomial

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Best to factor 5 out of the first 2 terms:

5x^2-7x+2=0 => 5(x^2 - [7/5]x) +2=0

Take half of [-7/5]: That'd be -7/10.

Square 7/10, add it to 5(x^2 - [7/5]x) +2=0 and then subtract it:


5(x^2 - [7/5]x)+ 49/100 - 49/100 ) +2=0

Then we have 5(x- [7/10] )^2 - 49/100 ) + 200/100 = 0

5(x-7/5)^2 - 245/100 + 200/100 = 0
5(x-7/5)^2 - 45/100 = 0
Dividing all terms by 5, x-7/5 = plus or minus sqrt( 9/100 )
x-7/5 = plus or minus 3/10

Then one root is x = 7/5 + 3/10, or x= 17/10 (answer #1)

The other is x = 7/5 - 3/10, or x = 14/10 - 3/10, or x = 7/10 (answer #2)

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