How do you solve by completing the square, leaving answers in simplest radical form: $x^2+ 3x- 2 = 0$ ?

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Answer 1 · 2015-04-03T00:42:56

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$x^2+3x-2=0$

$(x^2+3x color(white)("sssss"))-2=0$

$!/2$ of $3$ is $3/2$ . Square that to get $9/4$ add $9/4$ to complete the square and subtract to keep the equation balanced:

$(x^2+3x +9/4-9/4)-2=0$ Regoup to keep just the perfect square in the parentheses:

$(x^2+3x +9/4)-9/4-2=0$

Factor the perfect square (use the $3/2$ from before and " $-$ " like in the $+3x$ . Also simplify $-9/2-2$

$(x +3/2)^2-9/4-8/4= (x +3/2)^2-17/4= 0$

Solve $(x +3/2)^2-17/4= 0$ by "the Square Root Method"

$(x +3/2)^2-17/4= 0$

$(x +3/2)^2= 17/4$

$x +3/2 = +-sqrt(17/4) = +-sqrt17/sqrt4$

$x +3/2 = +- sqrt(17)/2$

$x = -3/2 +- sqrt(17)/2 = (-3 +- sqrt(17))/2$

(Use whichever form your teacher prefers -- one fraction or two.)

How do you solve by completing the square, leaving answers in simplest radical form: x^2+ 3x- 2 = 0x^2+ 3x- 2 = 0?