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41 votes
41 votes
Solve 2x^2+3x-1=0 by completing the square

User Syed Nasir Abbas
by
2.5k points

1 Answer

20 votes
20 votes

Answer:

x=(-3-√17)/4

x=(-3+√17)/4

Explanation:

2x^2+3x-1=0

it can rewrite as,

x^2+3x/2-1/2=0

now adding and subtracting (3/4)^2 , in R.H.S,

x^2+3x/2+9/16-9/16-1/2=0

(x+3/4)^2-17/16=0

(x+3/4)^2-((√17)/4)^2=0

(x+3/4+√17/4)(x+3/4-√17/4)=0

x=(-3-√17)/4

x=(-3+√17)/4

User Maluen
by
2.5k points
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