# Complete the square. 2x^2 - 3x - 7 = 0

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### 1 Answer

You need to keep the first two terms, that contain `x^2` and `x` and since between these two terms there exists minus sign, you need to use the following formula, such that:

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

Equating `a^2 = 2x^2` and `3x = 2ab` yields:

`a^2 = 2x^2 => a = sqrt2*x`

Replacing `sqrt2*x` for a in equation `3x = 2ab` yields:

`3x = 2sqrt2*x*b => 3 = 2sqrt2*b => b = 3/(2sqrt2) => b = 3sqrt2/4 => b^2 = 18/16`

You need to complete the square `a^2 - 2ab = 2x^2 - 3x` adding and then subtracting `b^2 = 18/16` , such that:

`(2x^2 - 3x + 18/16) - 18/16 - 7 = 0`

`(2x^2 - 3x + 18/16) - 130/16 = 0 => (sqrt2*x - 3sqrt2/4)^2 - 130/16 = 0`

**Hence, completing the square, under the given conditions, yields `2x^2 - 3x - 7 = (sqrt2*x - 3sqrt2/4)^2 - 130/16 = 0` .**