How do you solve the equation 2x^2+3x-5=0 by completing the square?

1 Answer
Tony B
Dec 21, 2016

x=-5/2 " and x=1

Explanation:

Write as 2(x^2+3/2x)-5+k=0.......Equation(1)

Where k is a constant of correction that compensates for the error introduce whilst manipulating the equation.

Take the power outside the bracket.

2(x+3/2x)^2-5+k=0

Remove the x from 3/2x

2(x+3/2)^2-5+k=0

Halve the 3/2

2(x+3/4)^2-5+k=0

color(red)("The error comes from")

color(red)(2)(x+color(red)(3/4))^(color(red)(2))-5+k=0...Equation(1_a)
.................................................................................................................
in that we have color(red)(2xx(3/4)^2)+k=0 larr" building the error correction"

=>k=-(2xx9/16) = -18/16 = -9/8
.....................................................................................................................

=>2(x+3/4)^2-5-9/8=0

=>color(blue)(2(x+3/4)^2-49/8=0)....................Equation(1_b)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(x+3/4)^2=49/16

x+3/4=+-sqrt(49/16)" "=" "+-7/4

x=+-7/4-3/4

x=-10/4 " and "1

color(red)(x=-5/2 " and "1)

Tony B

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