Question 202130
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2 - x = 21]


Standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2 - x - 21 = 0]


Start with:  *[tex \Large (2x\ \ \ )(x\ \ \ )] because that is the only way to get *[tex \Large 2x^2] as the high order term.  Next we know that one of the signs has to be positive and the other negative so that you end up with a -21.  But which is which?  We know the factors of 21 are 7 and 3, and we also know that the lead coefficient, 2, has to be multiplied by one of those to get the middle term -- Ah ha! 2 times 3 is 6 plus -7 is -1, the value of the given 1st degree term coefficient, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x - 7)(x + 3) = 0].


Use the Zero Product Rule (*[tex \Large a\cdot b = 0 \ \ \Leftrightarrow\ \ a = 0 \text{ or }b = 0]), so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x - 7 = 0 \ \ \Leftrightarrow\ \ x = \frac{7}{2}] or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 3 = 0 \ \ \Leftrightarrow\ \ x = -3]



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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 - 12x = 9]


Step 1, the lead coefficient must be 1, so divide through by the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 4x = 3]


Step 2, put the constant term on the right hand side.  This is already done for you, so go on to...


Step 3, divide the 1st degree coefficient by 2 and square the result: 4 divided by 2 is 2.  2 squared is 4.  Add this result to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 4x + 4 = 7]


Step 4, factor the resulting perfect square trinomial on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 2)^2 = 7]


Step 5, take the square root of both sides remembering to consider both the positive and negative root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 2 = \pm\sqrt{7}]


Step 6, finish solving for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 2\pm\sqrt{7}]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 + 11x - 4 = 0]


For all 2nd degree polynomial equations of the form *[tex \Large ax^2 + bx + c = 0], the roots are given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


Just do the substitutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(11) \pm sqrt{(11)^2 - 4(3)(-4)}}{2(3)} ]


You can do your own arithmetic.

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(6-x) = x(x+5)]


This one has to be put into standard form first.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12 - 2x = x^2 + 5x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x^2 -5x - 2x + 12 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 7x - 12 = 0]


Now just do the exact same thing we did in the previous problem except that now *[tex \Large a = 1], *[tex \Large b = 7], and *[tex \Large c = -12]


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The given roots are TMI; you don't need them.  All you need is the equation *[tex \Large 3x^2 + 11x - 4 = 0], *[tex \Large a] is the lead coefficient,*[tex \Large b] is the coefficient on the 1st degree term, and *[tex \Large c] is the constant term.




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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