Question

Solve by the quadratic formula. Simplify your answer.
x^2-3x-1=0

Answer 1

I would use x as a

Answer 2

http://www.algebra.com/algebra/homework/quadratic/solve-quadratic-equation.solver

Answer 3

x^2-3x-1=0

Answer 4

so in ur equation
a = 1
b = -3
c = -1

Answer 5

solve for x using the formula attached above

Answer 6

To be general when it comes to solving quadratic equations of the form
\[\Large \color{red}ax^2 + \color{blue}bx + \color{green}c =0\]
We will use the method of 'completing the square'
First, we need to have the \(x^2\) alone, with no coefficient, so the first step is to divide everything by \(\color{red}a\)
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{green}c}{\color{red}a}=0\]
Next, we subtract \(\Large \frac{\color{green}c}{\color{red}a}\) from both sides to that all the terms without \(x\) go on the right side...
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\]


Now, to the actual completing of the square... recall that to complete the square given
\[\Large x^2 + \color{purple}px= k\]we take half of the coefficient of the lone \(x\) (the one with no exponent), in this case, \(\color{purple}p\), giving \(\Large \frac{\color{purple}p}{2}\) and then square it, yielding \(\Large \frac{\color{purple}p^2}{4}\) and add THAT to both sides of the equation...\[\Large x^2 +\color{purple}px+\frac{\color{purple}p^2}{4}=k+\frac{\color{purple}p^2}{4}\]such that the left side is now a perfect square...\[\Large \left(x+\frac{\color{purple}p}2\right)^2 = k+\frac{\color{purple}p^2}{4}\]


It so happens that in this case, our \(\color{purple}p\) value is \(\Large \frac{\color{blue}b}{\color{red}a}\). So half of that is \(\Large \frac{\color{blue}b}{2\color{red}a}\). Squaring this yields \(\LARGE \frac{\color{blue}b^2}{4\color{red}a^2}\) And this is now added to both sides of the equation \(\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\) giving us...

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2}{4\color{red}a^2}-\frac{\color{green}c}{\color{red}a}\]

At this point, we may want to simplify the right-hand side of the equation into a single fraction, this can be done using their LCD, which happens to be \(\large 4\color{red}a^2 \)

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]


Now, as intended, the left-hand side is now a perfect square, so...
\[\Large \left(x+\frac{\color{blue}b}{2\color{red}a}\right)^2=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]

Taking the square root of both sides yields...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\pm\sqrt{\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}}\]
Simplifying...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\frac{\pm\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

Bringing the term \(\Large \frac{\color{blue}b}{2\color{red}a}\) to the right-hand side by subtracting from both sides gives
\[\Large x=-\frac{\color{blue}b}{2\color{red}a}\pm \frac{\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

And finally, combining, since they have the same denominator, we get...
\[\Large x= \frac{-\color{blue}b\pm \sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]


And this is the ever-notorious quadratic formula ^_^

Answer 7

Why you are proving it lol @terenzreignz

Answer 8

I wrote this out a long time ago....and because I'm bored XD

Answer 9

And the funniest part is the asker has gone offline

Answer 10

Pity :D

Answer 11

It's so frustating when someone asks a question and when u type a freakin long answer they just go offline

Answer 12

I didn't actually type that question... (not today, anyway)
It was typed a long time ago, but I saved it in case I want to present it again...
so it's the asker's loss, really XD

Answer 13

err I meant type that *response*

Answer 14

k