Question 456324
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 90]


is not an equation, quadratic or otherwise.  Equations have to contain an equals sign and show that two things are equal to one another.  A quadratic equation in standard form has a quadratic trinomial equal to zero.  Presuming that is what you meant, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 90\ =\ 0]


Proceed as follows:


<i><b><u>Factoring</u></b></i>


The lead coefficient is 1, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ \ \ )(x\ \ \ )\ =\ 0]


The sign on the constant term is negative so the signs in the two binomial factors are opposite:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ )(x\ +\ )\ =\ 0]


Now we need two integers whose product is -90 and whose sum is -1.  How about -10 and 9?


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 10)(x\ +\ 9)\ =\ 0]


Finally use the Zero Product Rule and set each of the binomials equal to 0 and solve each one to get the two roots you expect whenever you solve a quadratic.


<i><b><u>Quadratic Formula</u></b></i>


The quadratic formula is the general solution to the general quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


namely:


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


In the case of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 90\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -1\ \ ], and *[tex \LARGE c\ =\ -90]


Just plug in the numbers and do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-1)\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(-90)}}{2(-1)}]


You can do your own arithmetic.


<i><b><u>Graphing</u></b></i>


{{{drawing(
500, 500, -20,20,-20,20,
grid(1),
graph(
500, 500, -20,20,-20,20,
x^2-x-90))}}}


Zeros of the quadratic function 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ x^2\ -\ x\ -\ 90]


aka roots of the quadratic equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 90\ =\ 0]


are the *[tex \Large x]-coordinates of the points where the graph of the function intersects the *[tex \Large x]-axis, namely at *[tex \Large (-9,0)] and *[tex \Large (10,0)] as we expected from our previous work.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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