Question 71395
{{{3x^2 - 2x - 4 = 0}}}
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This equation does not factor nicely. So the best way to solve it is to use the quadratic 
formula. (The quadratic formula is derived by the method of completing the square, 
so it is a generalized form of the process of completing the square.)
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The quadratic formula says that for an expression of the quadratic form:
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{{{ax^2 + bx + c = 0}}}
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The value(s) of x is given by the equation:
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{{{x = -(b)/2a +- (sqrt(b^2 - 4*a*c))/2a}}}
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By comparing the numeric values given in the problem with the letters in the quadratic 
form you can see that a (the coefficient of x^2) is 3, b (the coefficient of x is -2, 
and c (the constant) is -4.  Now all you have to do is substitute those values into the 
equation for x.  When you do you get:
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{{{x = -(-2)/(2*3) +- (sqrt((-2)^2 - 4*3*(-4)))/(2*3)}}}
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The first stage of simplification results in:
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{{{x = 2/6 +- (sqrt(4 + 48))/6}}}
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Notice that the square root term becomes {{{sqrt(52)}}}. This equals {{{sqrt(13*4)}}} 
which furthermore can be written as {{{sqrt(13) * sqrt(4)}}}. But the square root of 
4 is 2, so the final reduction is {{{2*sqrt(13)}}}. Put this into the equation as a 
replacement for {{{sqrt(4+48)}}} to get:
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{{{x = 2/6 +- (2*sqrt(13))/6}}}
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The six is a common denominator so you can put both numerators over this common denominator 
to get:
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{{{x = (2 +- (2*sqrt(13)))/6}}}
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And since 2 is a factor of both terms in the numerator it can be made a multiplier 
of the two numerator terms as follows:
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{{{x = (2*(1 +- sqrt(13)))/6}}}
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Finally, the multiplier of two in the numerator and the denominator have a common 
factor of 2.  Cancel it to make the final answer:
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{{{x = ((1 +- sqrt(13)))/3}}}
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These are the two values for x.
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{{{x = ((1 + sqrt(13)))/3}}} and {{{x = ((1 - sqrt(13)))/3}}}
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Hope this helps you to see how to work this equation.  Make sure that you work through 
each step so you can see how to identify a, b, and c in a quadratic equation, and then 
how to work the numbers. This method is useful because it is a general method that can 
be used to evaluate and solve quadratic equations, some significantly more complex 
than this one.