If the equations are simultaneous and if and then:
Solve this equation.
Not sure if I am quite getting this correct?
y = 2x - 3
y = -x^2 + 4x +5
I am trying to solve the above and compare the results to a graph I plotted with a straight line and a parabola. I am looking at where the graph intersects and at the moment there seems to be slight differneces between my values from the graph and the above.
2x - 3
-x^2 + 4x +5
2x - 2 = x - 1
2 2
2(-1) - 3 = -2 -3 = -5
4(-1) + 5 = -4 + 5 = 1
(x + 1) (x - 5)
x = - 1 or x = + 5
These x values conside with previous work I have done and don't seem wrong, please advise.
By comparison to the graph(s), the straight line graph has points (0, 1.5) and the parabola has points (0, 5).
so somewhere I sem to be not quite right?
If the two equations are equal to y then they're equal to each other. Do you understand that?
Can you solve it? It's just a quadratic equation you've to solve. Put all the terms to one side and use the quadratic formula or factore.
I don't seem to get it, this is what I have tried?
2x - 3 = -x^2 + 4x + 5
2x = -x^2 + 4x + 5 + 3
2x = -x^2 + 4x + 8
then
2x + x^2 + 4x + 8
6x + x^2 + 8 = y
Using Ax^2 + BX + c = 0
a = 6, b = 1, c = 8
using the quadratic formula
- b + or - square root b^2 - 4ac / 2a
- 1 + or - square root 1^2 - 4 x 6 x 8 / 2 x 6
= square root 1 - 192
square root of 191 = 13.82 which is divided by 12
= 1.15 + 1 = 2.15 or
1.15 - 1 = 0.15
I believe neither of the above roots are correct, the x- values I think should be -1 and 5 as previously?
hi David Green
you got the coefficients a,b and c wrong:
6x + x^2 + 8 = 0
we can write this as:
x^2 + 6x + 8 = 0
the coefficient with x^2 is 1,so a=1
the coefficient with x is 6,so b=6
the last coefficient is 8,so c=8.
can you solve now?
You're welcome!
Take a look here for the graph:
2x-3=-x^2+4x+5 - Wolfram|Alpha