Thread: How 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?

There is no simple "algebraic" method. You say you aren't supposed to solve it "graphically". Is some other numerical method allowed?

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?

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?

For Nonlinear equations like yours you can use secant method.

Explanation of Secant Method:

Suppose there's a nonlinear function $\displaystyle y = f(x)$ in the image given below:



(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: $\displaystyle x = -3.16715$

pn-1	pn-2	f(pn-1)	f(pn-2)	pn
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.