Completing the square part two

Quadratic equations Completing the square part two

We have seen how we can complete the square for equations with a positive $x$ term like $x^2 + 6x + 2 = 0$ but what about an equation where the $x$ term is negative, like $x^2 - 6x + 2 = 0?$

We can solve this in exactly the same way as we have done when the $x$ term is positive. We still take half of the $x$ coefficient, add it t0 $x$ and square it. It is just that the coefficient of the $x$ term is negative.

$x^2 - 6x + 2 = 0$

$(x - 3)^2 - 9 + 2 = 0$

$(x - 3)^2 - 7 = 0$

$(x - 3)^2 = 7$

$sqrt{(x - 3)^2} = pmsqrt{7}$

$x - 3 = pmsqrt{7}$

$x = 3pmsqrt{7}$

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