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anonymous
 3 years ago
x^2 + x – 4 = 0 How would this be solved by graphing?
anonymous
 3 years ago
x^2 + x – 4 = 0 How would this be solved by graphing?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0quadratic equation, do you know the formula?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. \[x =\frac{ b \pm \sqrt{b^2  4ac} }{2a}\] But I need the vertex, domain and range, x and y intercepts, and the axis of symmetry.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did that stuff years ago, hang on I will look it up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks. lol I just have trouble with solving the equation by graphing. :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the x and y intercepts are going to be the equation factored out in the quadratic formula...I'm sure you can do that right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the domain will be all x values not equal to the x and y intercepts that you get, and they can not make the denominator 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = b/2a

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0solved by graphing means that we graph the equation and look at where it crosses the x axis

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x = \frac{ 1 \pm \sqrt{1^2  4 (1) (4)} }{ 2 (1) }\] I get \[x = \frac{ 1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can leave it in those terms if you want, but type it in the calculator

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why are you both copying and pasting?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am not allowed to have a decimal answer though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha you give her wolfram , nice source and your a moderator...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0wolfram does nice graphs :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then make that happen

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0"solve by graphing" is prolly the least exact method of finding a solution. But it is a viable in order to see if you ar in the right ballpark. The wolf does go ahead and spit out al the properties such as the exact figures :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah its just apart of learning the different methods to solving quadratic equations. I also have to find the answer using factoring, quadratic formula, and completing the square.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you will get them and understand them, they are not bad at all

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0completing the square is nice; its the "proof" to the quadratic equation.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}\frac ca\] \[(x+\frac b{2a})^2=\frac{b^24ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^24ac}}{2a}\] \[x=\frac b{2a}\pm\frac{\sqrt{b^24ac}}{2a}\] \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0thats the a general quadratic setup, and by running thru the "complete the square" process, we end up with the quadratic formula. completeing the square is the proof to the quadratic formula.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0since the completeing the square and the quadratic formula are 2 sides of the same coin.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How do I write x^2 + x  4 = 0 in standard form?
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