FAN AND MEDAL!
How do you solve -x^2+5x+6=0 ?

- anonymous

FAN AND MEDAL!
How do you solve -x^2+5x+6=0 ?

- schrodinger

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- anonymous

plz help

- zmudz

quadratic formula should do the trick, or you could factor into \(-(x-6)(x+1)\) and solve for x that way

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## More answers

- anonymous

@zmudz i did this but its wrong
-x^2+5x+6=0
(-x^2-x)+(6x+6)=0
-x(x+1)+6(x+1)=0
(-x+6)(x+1)=0
-x+6=0
-x=-6
x=6
x+1=0
x=-1

- zmudz

Ah, double check your math for when you solved x=-1

- anonymous

no they are both wrong i used a graphing calculator and the zeroes are x=0 and x=-1.2

- anonymous

it wants me to solve this system algebraically

- anonymous

Lol your answers x=6 and x=1 are both correct.

- anonymous

@phi i don't know what i did wrong in solving this

- anonymous

If they want you to solve it algebraically, then why did you use a graphing calculator? :D

- phi

I would first multiply both sides and all terms by -1
(because I don't like -x^2)
what do you get ?

- anonymous

because the second part is for me to graph it to prove my answers

- anonymous

x^2-5x-6=0 @phi

- phi

now try factoring it

- anonymous

How I wish we were allowed to use a graphing calculator when we graph something. Do you know how to plot points like this? |dw:1439243109664:dw|

- phi

but it looks like you did it correctly up top
x= 6 and x=-1

- anonymous

(x^2-2x)-(3x-6)=0
x(x-2)-3(x-2)=0
(x-3)(x-2)=0
x=3 and x=2

- anonymous

wait i messed up on the sign there

- anonymous

ok I'm confused @phi

- anonymous

This is if you graph it

##### 1 Attachment

- phi

(x^2 -6x) + (x-6) = 0

- phi

ok, so
\[ x^2 = 5x+6 \\ x^2 -5x-6=0\\ (x-6)(x+1)= 0 \\ x= 6 ; x= -1 \]
y= 5x+6 so at x= -1, y= 1
at x= 6 , y= 36
here is the graph with the first solution showing

##### 1 Attachment

- phi

the solution is where the two curves cross (intersect)
it is true when we are given a single curve (or line), we are often interested in where it crosses the x-axis (the x value is called the zero)
but not in this case.

- anonymous

and thats all?

- phi

solving two equations is the same as asking "what point lies on both curves"

- phi

there are two solutions
-1,1
and (6,36)

##### 1 Attachment

- anonymous

ok thank you!

- phi

you had the right idea except for the final step...
once you found x= -1 or x=6
you find the y value (using either equation)
and the (x,y) pairs are where the two curves intersect. i.e. are the "solutions"

- anonymous

oh ok thank you i have to go :)

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