Hey mathhelpforum.com, this is my first post! And I need some help with math!

The equation is 3^x+5=x^2-5, we aren't supposed to be able to solve them yet non-graphically, so I wonder if it is even possible, if so, how?

- April 12th 2013, 12:34 PMmoskowfreak1111How to find points of intersection of a parabola and an exponential w/o graphing?
Hey mathhelpforum.com, this is my first post! And I need some help with math!

The equation is 3^x+5=x^2-5, we aren't supposed to be able to solve them yet non-graphically, so I wonder if it is even possible, if so, how? - April 12th 2013, 04:34 PMHallsofIvyRe: How to find points of intersection of a parabola and an exponential w/o graphing?
There is no simple "algebraic" method. You say you aren't supposed to solve it "graphically". Is some other numerical method allowed?

- April 12th 2013, 06:53 PMmoskowfreak1111Re: How to find points of intersection of a parabola and an exponential w/o graphing?
Well, we can, and that's the only way to solve it is graphically my teacher says, but I don't want to, I want to actually learn how to do it, what would be the non-simple ways to do it?

- May 15th 2013, 01:42 PMx3bnmRe: How to find points of intersection of a parabola and an exponential w/o graphing?Quote:

The equation is 3^x+5=x^2-5, we aren't supposed to be able to solve them yet non-graphically, so I wonder if it is even possible, if so, how?

**Explanation of Secant Method:**

Suppose there's a nonlinear function in the image given below:

Attachment 28376

**(a)**Now take any two values of along the axis Say the first one is and the second one is

**(b)**plug the value of and into and solve for .

**(c)**You will get two points. They are and

**(d)**draw a secant line that goes through and

That secant line is

intersects axis at

**(e)**If you look carefully we can now write:

**(f)**Plug the values of , , and and you will get the value

**(g)**Now plug the value of into as the value of and solve for

you'll get the point

**(h)**A secant line can be drawn from to the point which is .

intersects axis at

**(i)**Again we can write:

Now plug in , , and and solve for

**(j)**Again plug this value of into and find

**(k)**Now continue this process iteratively until a value of is reached that will make closer to zero.

And that's it that's the root of the equation.

We can write Equation(1) as:

__Solving your question:__

You want to find the root of the equation is:

So the function is:

**(1)**Take any two values along the axis. Say the first point and

And:

**(2)**Now plug in into Equation(2) and solve for

You'll get

**(3)**Now let and

And do step (1) to (3) then again and again until plugging in the new found in each step as value of into gets reasonably closer to zero.

After 10 iterations I got the root of . And it's

If I plug this value in I get which is reasonably close.

So the approximate answer is:

**p**_{n-1}

**p**_{n-2}**f(p**_{n-1})**f(p**_{n-2})**p**_{n}3 2 28 15 0.84615 0.84615 3 11.81750 28 -0.72673 -0.72673 0.84615 9.92191 11.81750 -8.95951 -8.95951 -0.72673 -70.27277 9.92191 -1.74531 -1.74531 -8.95951 7.10088 -70.27277 -2.40739 -2.40739 -1.74531 4.27549 7.10088 -3.40928 -3.40928 -2.40739 -1.59957 4.27559 -3.13651 -3.13651 -3.40928 0.19418 -1.59957 -3.16604 -3.16604 -3.13651 0.00705 0.19418 -3.16715 -3.16715

-3.16604 -0.00002 0.00705 -3.16715

Hope this answers your question.