Question

how i solve this f(x)=0.3582*x^2 - 3.3833 * x + 9.0748 by matlab

Answer 1

how i solve this f(x)=0.3582*x^2 - 3.3833 * x + 9.0748

When asked to solve f(x) = 0, that means find the value(s) of x that, when substituted into the expression for f(x) will make it equal to zero.

So, to find f(x) = 0, then we set 0.3582*x^2 - 3.3833 * x + 9.0748 = 0.

This is a quadratic equation, of the form ax^2 + bx + c = 0, with a = 0.3582, b = - 3.3833 and c = 9.0748.

We use the quadratic formula to solve quadratic equations. This is,

x = {-b +/- sqrt(b^2 - 4ac)} / (2a)

substituting for the values given for a, b and c,

x = {3.3833 +/- sqrt((-3.3833)^2 - 4*(0.3582)*(9.0748))} / (2*0.3582)

x = {3.3833 +/- sqrt(11.4467 - 13.0024)} / (0.7164)

x = {3.3833 +/- sqrt(-1.55565)} / (0.7164)

Since the discriminant is negative (-1.55565), that means that we have no real values for x.

The expression, 0.3582*x^2 - 3.3833 * x + 9.0748, is a polynomial of 2nd degree. If you were to plot this curve you would get a parabola (U-shaped upwards) and its vertex would be above the x-axis. That means the curve never crosses the x-axis, which is why there is no (real) solution.

Our solution(s) then are complex numbers, made up of real and imaginary parts.

We write the square root part as:  sqrt(-1.55565) = sqrt(1.55565) * sqrt(-1) = sqrt(1.55565)}*i,

where i = sqrt(-1).

So now we have,

x = {3.3833 +/- sqrt(1.55565)*i} / (0.7164)

x = {3.3833 +/- 1.24726*i} / (0.7164)

x = 4.72264 +/- 1.741*i

Comments

When I first "answered" this question, I thought you were signing off your question with your username, matlab.

Rod spotted that you actually wanted to know how to use mMatlab to solve the problem, and his code for Matlab is spot on.

The only thing I can add is the solve function to use and that is roots(p) where p is the polynomial function that needs to be solved.

The full code to solve the problem would be.

>> p=[0.3582, -3.3833, 9.0748];
>> x=0:0.1:8;

>> f=polyval(p,x);
>> plot(x,f)
>> grid on
>> roots(p)

ans =

   4.7226 + 1.7410i
   4.7226 - 1.7410i

>>

Answer 2

Unfortunately, I do not have MATLAB, but here is a graph of the function in which you can see the shape of the parabolic curve, the vertex (minimum value) and the y intercept. You can also see that, because the graph does not intercept the x axis, there are no real roots. If I were to guess at the Matlab code, it would be something like:

% First, define the polynomial:

p = [ 0.3582, -3.3833, 9.0748 ];

% This range for x comfortably shows the graph and its key elements:

x = -2 : 0.1 : 8; 

% Plot the graph:

f = polyval (p ,x);

plot (x,f)

grid on