Question 329149
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Three ways to do this, two of them practical in this case.


Factoring:


Recognize that *[tex \LARGE -4\ \times\ 11\ =\ -44] and *[tex \LARGE -4\ +\ 11\ = 7].


Then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 7x\ -\ 44\ =\ (x\ -\ 4)(x\ +\ 11)\ =\ 0]


Then use the Zero Product Rule, namely *[tex \Large ab\ =\ 0] if and only if *[tex \Large a\ =\ 0] or *[tex \Large b\ =\ 0], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 4\ =\ 0\ \Leftrightarrow\ \ x\ =\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 11\ =\ 0\ \Leftrightarrow\ \ x\ =\ -11]


Or use the quadratic formula, which is the general solution to the quadratic equation *[tex \Large ax^2\ +\ bx\ +\ c\ =\ 0]:


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


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 7\ \ ], and *[tex \LARGE c\ =\ -44]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(7)\ \pm\ \sqrt{(7)^2\ -\ 4(1)(-44)}}{2(1)}]


I'll let you do your own arithmetic to determine that *[tex \Large x\ =\ 4] or *[tex \Large x\ =\ -11]


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