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

Answer 1

Ethan Adkins

In this manner:

#x^2+3x-18=0rArrx^2+3x+(3/2)^2-(3/2)^2-18=0rArr#

#(x^2+3x+9/4)-9/4-18=0rArr(x+3/2)^2=18+9/4rArr#

#(x+3/2)^2=(72+9)/4rArr(x+3/2)^2=81/4rArr#

#x+3/2=+-9/2rArrx=-3/2+-9/2rArr#

#x_1=-3/2-9/2=-12/2=-6#

#x_2=-3/2+9/2=6/2=3#.

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Answer 2

Eli Assouline

To solve the quadratic equation (x^2 + 3x - 18 = 0) by completing the square:

Move the constant term to the other side of the equation: (x^2 + 3x = 18)
Take half of the coefficient of (x), square it, and add it to both sides of the equation: (x^2 + 3x + \left(\frac{3}{2}\right)^2 = 18 + \left(\frac{3}{2}\right)^2) (x^2 + 3x + \frac{9}{4} = 18 + \frac{9}{4})
Factor the perfect square trinomial on the left side: ((x + \frac{3}{2})^2 = \frac{81}{4})
Take the square root of both sides: (x + \frac{3}{2} = \pm \frac{9}{2})
Solve for (x): (x = -\frac{3}{2} \pm \frac{9}{2})
Simplify: (x = -6) or (x = 3)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.