OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

quadratic equation, do you know the formula?

5 years ago
OpenStudy (anonymous):

yes. \[x =\frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}\] But I need the vertex, domain and range, x and y intercepts, and the axis of symmetry.

5 years ago
OpenStudy (anonymous):

I did that stuff years ago, hang on I will look it up

5 years ago
OpenStudy (anonymous):

Thanks. lol I just have trouble with solving the equation by graphing. :/

5 years ago
OpenStudy (anonymous):

the x and y intercepts are going to be the equation factored out in the quadratic formula...I'm sure you can do that right

5 years ago
OpenStudy (anonymous):

the 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

5 years ago
OpenStudy (anonymous):

The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = -b/2a

5 years ago
OpenStudy (amistre64):

solved by graphing means that we graph the equation and look at where it crosses the x axis

5 years ago
OpenStudy (anonymous):

\[x = \frac{ -1 \pm \sqrt{1^2 - 4 (1) (-4)} }{ 2 (1) }\] I get \[x = \frac{ -1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?

5 years ago
OpenStudy (anonymous):

you can leave it in those terms if you want, but type it in the calculator

5 years ago
OpenStudy (anonymous):

why are you both copying and pasting?

5 years ago
OpenStudy (anonymous):

I am not allowed to have a decimal answer though.

5 years ago
OpenStudy (anonymous):

haha you give her wolfram , nice source and your a moderator...

5 years ago
OpenStudy (amistre64):

wolfram does nice graphs :)

5 years ago
OpenStudy (amistre64):

to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method

5 years ago
OpenStudy (anonymous):

its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers

5 years ago
OpenStudy (anonymous):

Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.

5 years ago
OpenStudy (anonymous):

Then make that happen

5 years ago
OpenStudy (amistre64):

"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 :)

5 years ago
OpenStudy (anonymous):

Yeah 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.

5 years ago
OpenStudy (anonymous):

you will get them and understand them, they are not bad at all

5 years ago
OpenStudy (amistre64):

completing the square is nice; its the "proof" to the quadratic equation.

5 years ago
OpenStudy (amistre64):

\[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=-\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\] \[(x+\frac b{2a})^2=\frac{b^2-4ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=-\frac b{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

5 years ago
OpenStudy (anonymous):

Whats all that?! lol

5 years ago
OpenStudy (amistre64):

thats 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.

5 years ago
OpenStudy (amistre64):

since the completeing the square and the quadratic formula are 2 sides of the same coin.

5 years ago
OpenStudy (anonymous):

True

5 years ago
OpenStudy (anonymous):

How do I write x^2 + x - 4 = 0 in standard form?

5 years ago