Solve using Gauss Elimination method
X+2Y+Z=6
2X+Y+3Z=12
3X+2Y+4Z=17.
X+2Y+Z=6
2X+Y+3Z=12
3X+2Y+4Z=17.
Find the roots of the equation f(x) = x2+3x-3=0 using regula falsi method?
The regula falsi method is an iterative method of root finding that uses a linear approximation of the function between functional values of opposite sign known to bracket the root. It is also known as the secant method.
The first step is to find values of the independent variable for which values of the function have opposite sign. Descartes' Rule of Signs and the Rational Root Theorem can be helpful in this.
For your function
f(x) = x^2 + 3x - 3
we know that f(0) = -3 and f(1) = +1, so one of the roots lies in the interval (0, 1).
The second step is to perform the iteration. We can use as the iteration formula
x = x1 - (x2 - x1)/(y2 - y1)*y1
where x1 and x2 are chosen such that y1=f(x1) and y2=f(x2) have opposite signs and one of them is the most recent value of x.
We start with (x1, y1) = (0, -3) and (x2, y2) = (1, 1). Our next iteration is
x = 0 - (1-0)/(1-(-3))*(-3) = 0 - (-3/4) = 3/4
f(3/4) = 9/16 + 9/4 - 3 = -3/16
Now, we use (x1, y1) = (3/4, -3/16) replacing the previous value of negative sign. The next iteration is
x = 3/4 - (1-3/4)/(1-(-3/16))*(-3/16) = 15/19
f(15/19) = -3/361, still of negative sign. So we replace (x1, y1) once again. The next iteration is
x = 15/19 - (1-15/19)/(1-(-3/361))*(-3/361) = 72/91 ≈ 0.791209
f(72/91) = -3/8281 ≈ -.000362275, still of negative sign. Continuing, the next few iterations are (x, f(x))
{0.791284, -0.0000157815},
{0.791288, -6.87455*10^-7},
{0.791288, -2.99461*10^-8}
Thus, one of the roots is x = .791288. Note that we achieved 6 digit accuracy in 5 iterations on this problem. Dividing the polynomial by the factor (x-.791288), we get (x+3.79129). This tells us the other root is x = -3.79129.
We could iterate starting with the interval (0, -4) to get a similar result.
Actual roots are (-3±√21)/2 ≈ (-3.79129, 0.791288).
The first step is to find values of the independent variable for which values of the function have opposite sign. Descartes' Rule of Signs and the Rational Root Theorem can be helpful in this.
For your function
f(x) = x^2 + 3x - 3
we know that f(0) = -3 and f(1) = +1, so one of the roots lies in the interval (0, 1).
The second step is to perform the iteration. We can use as the iteration formula
x = x1 - (x2 - x1)/(y2 - y1)*y1
where x1 and x2 are chosen such that y1=f(x1) and y2=f(x2) have opposite signs and one of them is the most recent value of x.
We start with (x1, y1) = (0, -3) and (x2, y2) = (1, 1). Our next iteration is
x = 0 - (1-0)/(1-(-3))*(-3) = 0 - (-3/4) = 3/4
f(3/4) = 9/16 + 9/4 - 3 = -3/16
Now, we use (x1, y1) = (3/4, -3/16) replacing the previous value of negative sign. The next iteration is
x = 3/4 - (1-3/4)/(1-(-3/16))*(-3/16) = 15/19
f(15/19) = -3/361, still of negative sign. So we replace (x1, y1) once again. The next iteration is
x = 15/19 - (1-15/19)/(1-(-3/361))*(-3/361) = 72/91 ≈ 0.791209
f(72/91) = -3/8281 ≈ -.000362275, still of negative sign. Continuing, the next few iterations are (x, f(x))
{0.791284, -0.0000157815},
{0.791288, -6.87455*10^-7},
{0.791288, -2.99461*10^-8}
Thus, one of the roots is x = .791288. Note that we achieved 6 digit accuracy in 5 iterations on this problem. Dividing the polynomial by the factor (x-.791288), we get (x+3.79129). This tells us the other root is x = -3.79129.
We could iterate starting with the interval (0, -4) to get a similar result.
Actual roots are (-3±√21)/2 ≈ (-3.79129, 0.791288).
Well what ever is the method. The answer is very simple (a little factorisation)
x2+3x-3=0 , I belive is a second order quadratic equation, which
you can solev by using quadratic formual ior by completeing the square method.
The simple method is by completeing the square. Give it a try.
x2+3x-3=0 , I belive is a second order quadratic equation, which
you can solev by using quadratic formual ior by completeing the square method.
The simple method is by completeing the square. Give it a try.
