solve by completing the square2x^2+3x=20

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`2x^2+3x = 20`

`2x^2+3x-20 = 0`

`1/2(2x^2+3x-20) = 1/2xx0`

`x^2+(3/2)x-10 = 0`

 

We know that;

`(x+a)^2 = x^2+2ax+a^2 `

 

`(x+3/4)^2 = x^2+(3/2)x+9/16`

`(x+3/4)^2 -10 = x^2+(3/2)x+9/16-10`

 

`(x+3/4)^2 -10 -9/16 = x^2+(3/2)x-10`

 

We know that `x^2+(3/2)x-10 = 0`

 

`(x+3/4)^2 -10 -9/16 = 0`

 `(x+3/4)^2 = 10+9/16`

 `(x+3/4)^2 = 169/16`

 `(x+3/4) = +-sqrt(169/16)`

 `(x+3/4) = +-(13/4)`

         `x = -3/4+-13/4`

 

`x = (-3-13)/4 = -4`

`x = (-3+13)/4 = 5/2`

 

So the answers are x = -4 and x = 5/2

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to complete the square to the left side using the following formula such that:

`(a + b)^2 = a^2 + 2ab + b^2`

Comparing the left side to the expansion of the square you may identify that:

`2x^2 = a^2 => => a = xsqrt 2`

`3x = 2ab => 3x = 2*x*sqrt2*b => b = 3/(2sqrt2) => b = (3sqrt2)/4`

You may complete the square such that:

`2x^2 + 3x + ((3sqrt2)/4)^2 = 20 + ((3sqrt2)/4)^2`

`(sqrt 2*x + (3sqrt2)/4)^2 = 20 + 18/16`

`(sqrt 2*x + (3sqrt2)/4)^2 = (320 + 18)/16`

`(sqrt 2*x + (3sqrt2)/4)^2 = 338/16`

`sqrt 2*x + (3sqrt2)/4 = +-(sqrt338)/4`

`sqrt 2*x + (3sqrt2)/4 = +-(13sqrt2)/4`

`sqrt 2*x = (13sqrt2 - 3sqrt2)/4 => sqrt 2*x = (10sqrt2)/4`

`x_1 = 5/2`

`sqrt 2*x = (-13sqrt2 - 3sqrt2)/4 => sqrt 2*x = (-16sqrt2)/4`

`x_2 = -4`

Hence, evaluating the solutions to quadratic equation, completing the given square, yields `x_1 = 5/2` and `x_2 = -4` .

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