Solve 4x^2+2x-3=0 by completing the square

dkarolasz

Junior Member
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Jun 6, 2007
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Solve by completing the square

4x^2+2x-3=0

(2x+1/2)2-1/2-3=0

(2x+1/2)2=7/2

Did I get this right?
 

Denis

Senior Member
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Feb 17, 2004
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1,460
google "completing the square"
 

jwpaine

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Mar 10, 2007
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4x^2+2x-3 = 0

divide out by 4

x^2 + (1/2)x - (3/4) = 0
x^2 + (1/2)x = (3/4)

add ((1/2)b)^2 to both sides

x^2 + (1/2)x + (1/16) = (3/4) + (1/16)

simplify the right hand side, convert the left hand side to squared form

(x + (1/4))^2 = (13/16)

square root both sides

x + (1/4) = sqrt(13/16)

thus x = -(1/4) +/- sqrt(13/16)


See?
John.
 

galactus

Super Moderator
Staff member
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Sep 28, 2005
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When completing the square, it helps if you just know the general formula:

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}-c+\frac{b^{2}}{4a}\)
 

Mrspi

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Dec 17, 2005
Messages
2,128
dkarolasz said:
Solve by completing the square

4x^2+2x-3=0

(2x+1/2)2-1/2-3=0

(2x+1/2)2=7/2

Did I get this right?
I approach completing the square this way:

1) get the terms containing the variable on one side, and the constant term on the other side:

4x<SUP>2</SUP> + 2x = 3

2) Divide both sides by the coefficient of the square term:

x<SUP>2</SUP> + (1/2)x = 3/4

3) Multiply the coefficient of x by 1/2, square the result, and add the square to both sides of the equation. (1/2)*(1/2) is 1/4. (1/4)<SUP>2</SUP> is 1/16. Add 1/16 to both sides of the equation:

x<SUP>2</SUP> + (1/2)x + (1/16) = (3/4) + (1/16)

4) Write the left side as the square of a binomial, and do the arithmetic on the right side:

[ x + (1/4)]<SUP>2</SUP> = 13/16

5) Take the square root of both sides

x + (1/4) = +/- sqrt(13/16)

x + (1/4) = +/- sqrt(13)/4

6) Get x by itself....

x = (-1/4) +/- sqrt(13) / 4
 
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