How do you solve by completing the square: $x^{2} + 3 x - 18 = 0$ $x^2 + 3x - 18 = 0$ ?

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How do you solve by completing the square: $x^{2} + 3 x - 18 = 0$ $x^2 + 3x - 18 = 0$ ?

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Answer 1 · 2015-03-30T14:13:23

In this way:

$x^{2} + 3 x - 18 = 0 \Rightarrow x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} - 18 = 0 \Rightarrow$ $x^2+3x-18=0rArrx^2+3x+(3/2)^2-(3/2)^2-18=0rArr$

$(x^{2} + 3 x + \frac{9}{4}) - \frac{9}{4} - 18 = 0 \Rightarrow {(x + \frac{3}{2})}^{2} = 18 + \frac{9}{4} \Rightarrow$ $(x^2+3x+9/4)-9/4-18=0rArr(x+3/2)^2=18+9/4rArr$

${(x + \frac{3}{2})}^{2} = \frac{72 + 9}{4} \Rightarrow {(x + \frac{3}{2})}^{2} = \frac{81}{4} \Rightarrow$ $(x+3/2)^2=(72+9)/4rArr(x+3/2)^2=81/4rArr$

$x + \frac{3}{2} = \pm \frac{9}{2} \Rightarrow x = - \frac{3}{2} \pm \frac{9}{2} \Rightarrow$ $x+3/2=+-9/2rArrx=-3/2+-9/2rArr$

$x_{1} = - \frac{3}{2} - \frac{9}{2} = - \frac{12}{2} = - 6$ $x_1=-3/2-9/2=-12/2=-6$

$x_{2} = - \frac{3}{2} + \frac{9}{2} = \frac{6}{2} = 3$ $x_2=-3/2+9/2=6/2=3$ .

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How do you solve by completing the square: $x^{2} + 3 x - 18 = 0$ $x^2 + 3x - 18 = 0$ ?

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