The problem says to solve for 'x' and then graph the equation?

Question

The problem is x^2=5x+2.
The answer that I came up with is
x=5+/-sqrt 33/2.
The question is-how do I graph that? Or is there a separate formula used for this?

Answer 1

Since there's no 'y' in your equation, there's really only one variable, so technically a graph of the equation consists of drawing those two points on a number line. Your teacher might also mean:

Graph the polynomial y=x²-5x-2 (since it is the roots of this equation that are solutions to the first)
Graph y=x² and also graph y=5x+2 (the x-coordinates of the points where the two curves intersect are precisely the solutions of your equation -- plus this gives you a good picture of what's happening to both sides of the equation as the value of x changes).

But the main thing you want to do at this point is ask your teacher what she wants you to draw.

Answer 2

1. get everything on one side
x^2-5x+2=0
2. find the vertex by -b/2a which is the x in the vertex (x,y)
5/2(1)=5/2
3. substitute 0 for y and plug in 5/2 to find the y in the vertex
(5/2)^2-5(5/2)+2=y
25/4-25/2+2=y
-17/4=7
vertex=(5/2,-17/4)
4. plug in #'s for x and find the y, those are the points the graph shoud go in, find at least 1 or 2 for both sides of the vertex and graph it

Answer 3

Given: 2x-y=8 and x+3y=11 sparkling up graphically. For the line 2x-y=8 (say line AB); has coordinates: A(0,-8); B(4,0) go with x = 0 then y = -8 or :2x-y=8 or if x = 0 then 2(0)-y=8 or y = -8 go with y = 0 then x = 4 :2x-y = 8 or if y = 0 then 2x+0=8 or x = 4 Draw the graph for this lineAB. For the line x+3y=11 (say line CD): has coordinates: C(0,11/3); D(4,0) go with x = 0 then y = 11/3 or :x+3y=11 or if x = 0 then 0+3y=11 or y = 11/3 go with y = 0 then x = 4 :x+3y = 11 or if y = 0 then x+3(0)=11 or x = 11 Draw the graph for this line CD. anticipate O via fact the factor of interesection. locate the coordinates of factor O O(7,6) <= answer

Answer 4

x^2 = 5x + 2
x^2 - 5x - 2 = 0

for a graph, go to http://www.calculator.com/calcs/GCalc.html

type in x^2 - 5x - 2, click enter on the keyboard and this will graph it for you.

x-intercepts when graph, about a third to the right of x = 5, and just a third to the left of x = 0, with a y-intercept of y = -2, Vertex is at (2.5,-8.25)

Answer 5

x^2 = 5x + 2
x^2 - 5x - 2 = 0

Using a^2 + bx - c = 0 and the eq. (-b +/- (sqrt (b^2-4ac))) 1/2a
x = 5.37 and -0.37

so den dese are the pts on the x-axis
on the y-axis, x = 0 so x = -2
as since the co-efficient of x^2 is positive it is a curve with a minimum turning pt.
and den join the pts

Answer 6

i know that to graph it, you need to get everything to one side. and it'd be y=-x^2+5x+2. then graph that. if you can, get access to a graphing calculator - it makes things wayyy simpler! if not, good luck to ya, mate. (don't ask why i went pirate there. i don't know!)
as for the solving for x part, use quadratic formula --> (b^2 +/- sqr b^2 - 4ac)/2a......you'd plug in a=-1, b=5, and c=2 :)