Question 84422
Given:
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{{{2x^2+10x+11=0}}}
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Subtract 11 from both sides to get the constant on the right side of the equation. This results
in the equation becoming:
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{{{2x^2 + 10x = -11}}}
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You want the multiplier (coefficient) of the {{{x^2}}} term to be 1. To make that happen,
divide every term on both sides of the equation by the multiplier of the {{{x^2}}} which is 2.
When you do that division the equation becomes:
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{{{x^2 + 5x = -11/2}}}
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Next, divide the multiplier of the x term by 2. That is divide the 5 by 2 to get {{{5/2}}}.
Square that term to get {{{25/4}}}. Now add {{{25/4}}} to both sides of the equation.
You get:
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{{{x^2 + 5x + 25/4 = -11/2 + 25/4}}}
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The left side factors to:
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{{{(x+(5/2))^2 = -11/2 + 25/4}}}
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Multiply the {{{-11/2}}} by {{{2/2}}} to get {{{-22/4}}} which now has a common denominator
with the other term on the right side. Substitute this into the equation to get:
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{{{(x+(5/2))^2 = (-22/4)+(25/4)}}}
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The right side combines to give:
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{{{(x+(5/2))^2 = 3/4}}}
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Now take the square root of both sides to get:
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{{{x + (5/2) = sqrt(3/4)}}}
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Note that when you take the square root of the right side, the square root is preceded by
both a + and a - sign.  This shows in the next step. Look carefully for the two signs
between the two terms on the right side.
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Subtract {{{5/2}}} from both sides and also transform the square root of 3/4 into the
square root of 3 divided by the square root of 4. When you do that you will get:
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{{{ x = -5/2 +- sqrt(3/4) = -5/2 +- sqrt(3)/sqrt(4)}}}
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The square root of 4 is 2 and when this is substituted the equation becomes:
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{{{x = -5/2 +- sqrt(3)/2}}}
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Hope this helps you to understand the process of completing the square to solve a quadratic
equation.