Question 837931
First of all, these are not exponential functions (which have variables in the exponents). These are polynomial functions. (I have changed the category to the appropriate one.)<br>
If P(x) = Q(x) then
{{{x^3-3x^2+2= x^2-6x+11}}}
Now we solve for x. To solve polynomials like this we want one side to be zero. Subtracting the entire right side from both sides we get:
{{{x^2-4x^2+6x-9=0}}}
And now we factor. The GCF is 1, this does not fit any of the factoring patterns and it is not a trinomial. And since I don't see how to use factoring by grouping, it appears that we must resort to trial and error of the possible rational roots.<br>
The possible rational roots of any polynomial are all the ratios, positive and negative, which can be formed using a factor of the constant term (at the end) in the numerator and a factor of the leading coefficient (at the front) in the denominator. With a constant term of 9 (whose factors are 1, 3 and 9) and a leading coefficient of 1, our possible rational roots are:
<u>+</u>1/1, <u>+</u>3/1 and <u>+</u>9/1
which simplify to:
<u>+</u>1, <u>+</u>3 and <u>+</u>9<br>
We can check 1 (and perhaps -1) with mental math. Neither one works here. SO we'll try 3. Synthetic division is often used when mental math is not practical:
<pre>
3  |   1   -4   6   -9
----        3  -3    9
      -----------------
       1   -1   3    0
</pre>And we have a root! The remainder, in the lower right corner, is zero. This means that 3 is a root (and that (x-3) is a factor) of {{{x^3-4x^2+6x-9}}} Not only that, the rest of the bottom row tells us the other factor. The "1 -1 3" translates into {{{x^2-x+3}}}.<br>
This other factor is a quadratic. It does not factor but we can use the quadratic formula:
{{{x = (-(-1)+-sqrt((-1)^2-4(1)(3)))/2(1)}}}
Simplifying...
{{{x = (1+-sqrt(1-4(1)(3)))/2}}}
{{{x = (1+-sqrt(1-12))/2}}}
{{{x = (1+-sqrt(-11))/2}}}
{{{x = (1+-sqrt(-1*11))/2}}}
{{{x = (1+-sqrt(-1)*sqrt(11))/2}}}
{{{x = (1+- i*sqrt(11))/2}}}
which is short for:
{{{x = (1+ i*sqrt(11))/2}}} or {{{x = (1- i*sqrt(11))/2}}}
or (in standard a + bi form):
{{{x = 1/2+ (sqrt(11)/2)i}}} or {{{x = 1/2+ (-sqrt(11)/2)i}}}<br>
So there are three values of x that make P(x) = Q(x): 3, {{{1/2+ (sqrt(11)/2)i}}}, {{{1/2+ (-sqrt(11)/2)i}}}<br>