How do you solve using completing the square method $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ ?

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How do you solve using completing the square method $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ ?

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Answer 1 · 2016-04-06T05:15:37

$x = - 3$ $x= -3$
$x = 6$ $x= 6$

Explanation:

Note: The goal of completing the square is to create a perfect trinomial in the form of

$(a^{2} + 2 a b + b^{2}) = {(a + b)}^{2}$ $(a^2 +2ab+b^2) = (a+b)^2$ $or$ $" " " or$
$(a^{2} - 2 a b + b^{2}) = {(a - b)}^{2}$ $(a^2 -2ab+b^2) = (a-b)^2$

Given:

$x^{2} - 3 x = 18$ $x^2 - 3x = 18$

Step 1: The leading coefficient is 1, we can proceed to the next step of completing the square

Step 2: The third term is , ${[(\frac{1}{2}) (- 3)]}^{2} = \frac{9}{4}$ $[(1/2)(-3)]^2 = 9/4$

$x^{2} - 3 x + {[(\frac{1}{2}) (- 3)]}^{2} = 18 + {[(\frac{1}{2}) (- 3)]}^{2}$ $x^2 - 3x + [(1/2)(-3)]^2 = 18 + [(1/2)(-3)]^2$

$x^{2} - 3 x + \frac{9}{4} = 18 + \frac{9}{4}$ $x^2 - 3x +color(red)(9/4) = 18 + color(red)(9/4)$

Step 3: Rewrite as a perfect trinomial on the left and simplify the right side of the equation

$x^{2} - 3 x + {(\frac{3}{2})}^{2} = \frac{18}{1} + \frac{9}{4}$ $x^2 -3x +(3/2)^2 = 18/1 + 9/4$

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

${(x - \frac{3}{2})}^{2} = \frac{81}{4}$ $(x-3/2)^2 = 81/4$

Step 4: Solve the equation by taking the square root of both side
remember $\sqrt{x^{2}} = \pm x$ $sqrt(x^2) = +-x$

$\sqrt{{(x - \frac{3}{2})}^{2}} = \sqrt{\frac{81}{4}}$ $sqrt((x-3/2)^2) = sqrt(81/4)$

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

Step 5: Then solve for $x$ $x$

$x = - \frac{9}{2} + \frac{3}{2}$ $x = -9/2 + 3/2$ $or$ $" " or "$ $x = \frac{9}{2} + \frac{3}{2}$ $x= 9/2 +3/2$

So $x = - \frac{6}{2} = - 3$ $" " x= -6/2 = -3$

or $x = \frac{12}{2} = 6$ $" " x= 12/2 = 6$

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How do you solve using completing the square method $x^{2} - 3 x = 18$ $x^2 - 3x = 18$ ?

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Explanation:

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