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?

- Apr 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? - Apr 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?

- Apr 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 $\displaystyle y = f(x)$ in the image given below:

Attachment 28376

**(a)**Now take any two values of $\displaystyle x$ along the $\displaystyle X$ axis Say the first one is $\displaystyle p_0$ and the second one is $\displaystyle p_1$

**(b)**plug the value of $\displaystyle p_0$ and $\displaystyle p_1$ into $\displaystyle y = f(x)$ and solve for $\displaystyle y$.

**(c)**You will get two points. They are $\displaystyle (p_0, f(p_0))$ and $\displaystyle (p_1, f(p_1))$

**(d)**draw a secant line that goes through $\displaystyle (p_0, f(p_0))$ and $\displaystyle (p_1, f(p_1))$

That secant line is $\displaystyle L1$

$\displaystyle L1$ intersects $\displaystyle X$ axis at $\displaystyle p_2$

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

$\displaystyle \frac{f(p_1)}{p_1 - p_2} = \frac{f(p_1)-f(p_0)}{p_1 - p_0}......[\text{Equation (1)}]$

**(f)**Plug the values of $\displaystyle p_0$, $\displaystyle f(p_0)$, $\displaystyle p_1$ and $\displaystyle f(p_1)$ and you will get the value $\displaystyle p_2$

**(g)**Now plug the value of $\displaystyle p_2$ into $\displaystyle y = f(x)$ as the value of $\displaystyle x$ and solve for $\displaystyle y$

you'll get the point $\displaystyle (p_2, f(p_2))$

**(h)**A secant line can be drawn from $\displaystyle (p_1, f(p_1))$ to the point $\displaystyle (p_2, f(p_2))$ which is $\displaystyle L2$.

$\displaystyle L2$ intersects $\displaystyle X$ axis at $\displaystyle p_3$

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

$\displaystyle \frac{f(p_2)}{p_2 - p_3} = \frac{f(p_2)-f(p_1)}{p_2 - p_1}$

Now plug in $\displaystyle p_1$, $\displaystyle f(p_1)$, $\displaystyle p_2$ and $\displaystyle f(p_2)$ and solve for $\displaystyle p_3$

**(j)**Again plug this value of $\displaystyle x = p_3$ into $\displaystyle y =f(x)$ and find $\displaystyle f(p_3)$

**(k)**Now continue this process iteratively until a value of $\displaystyle x$ is reached that will make $\displaystyle y = f(x)$ closer to zero.

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

We can write Equation(1) as:

$\displaystyle \frac{f(p_{n-1})}{p_{n-1} - p_n} = \frac{f(p_{n-1})-f(p_{n-2})}{p_{n-1} - p_{n-2}}......[\text{Equation (2)}]$

__Solving your question:__

You want to find the root of the equation is:

$\displaystyle 3^x - x^2 + 10 = 0$

So the function is:

$\displaystyle f(x) = 3^x - x^2 + 10.......(3)$

**(1)**Take any two $\displaystyle x$ values along the $\displaystyle X$ axis. Say the first point $\displaystyle p_{n-2} = 2$ and $\displaystyle p_{n-1} = 3$

$\displaystyle f(p_{n-2}) = f(2) = 15.....\text{[by plugging in } x = 2 \text{into function labeled as (3)]}$

And:

$\displaystyle f(p_{n-1}) = f(3) = 28$

**(2)**Now plug in $\displaystyle p_{n-1}, f(p_{n-1}), p_{n-2}, f(p_{n-2})$ into Equation(2) and solve for $\displaystyle p_{n}$

You'll get $\displaystyle p_n = 0.84615$

**(3)**Now let $\displaystyle p_{n-2} = p_{n-1} = 3$ and $\displaystyle p_{n-1} = p_n = 0.84615$

And do step (1) to (3) then again and again until plugging in the new found $\displaystyle p_n$ in each step as value of $\displaystyle x$ into $\displaystyle y = 3^x - x^2 + 10$ gets reasonably closer to zero.

After 10 iterations I got the root of $\displaystyle y = 3^x - x^2 + 10$. And it's $\displaystyle x = -3.16715$

If I plug this value in $\displaystyle y = 3^x - x^2 + 10$ I get $\displaystyle -0.00002$ which is reasonably close.

So the approximate answer is: $\displaystyle x = -3.16715$

**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.