Question

How do you solve `2x^2 + 8x + 7 = 0`?

Answer 1

`x = -2+-sqrt(2)/2`

Explanation:

Completing the Square

Multiply by `2` to make the leading term into a perfect square, complete the square, then use the difference of square identity:

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

with `a=(x+2)` and `b=sqrt(2)` as follows:

`0 = 4x^2+16x+14`

`=(2x+4)^2-16+14`

`=(2x+4)^2-(sqrt(2))^2`

`=((2x+4)-sqrt(2))((2x+4)+sqrt(2))`

`=(2x+4-sqrt(2))(2x+4+sqrt(2))`

`=(2(x+2-sqrt(2)/2))(2(x+2+sqrt(2)/2))`

`=4(x+2-sqrt(2)/2)(x+2+sqrt(2)/2)`

Hence:

`x = -2+-sqrt(2)/2`

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Quadratic Formula

`2x^2+8x+7` is in the form `ax^2+bx+c` with `a=2`, `b=8` and `c=7`.

This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`=(-8+-sqrt(8^2-(4*2*7)))/(2*2)`

`=(-8+-sqrt(64-56))/4`

`=(-8+-sqrt(8))/4`

`=(-8+-sqrt(2^2*2))/4`

`=(-8+-2sqrt(2))/4`

`=-2+-sqrt(2)/2`