I'm not familiar with the "bisection method" to find the roots of 2x2-5x+1 = 0 but by completing the square or using the quadratic equation formula you'll find that the solution is:
x = (5 + or - the square root of 17) over 4
Hope that helps.
Square roots are computed using the Babylonian method, calculators, Newton's method, or the Rough estimation method. * * * * * Or the Newton-Raphson method.
It has roots x = 2.618 and x = 0.38197
No real roots
The roots are: x = -5 and x = -9
There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2
The bisection method is simpler to implement and guarantees convergence to a root if one exists within the initial interval, but it can be slower as it always halves the interval. In contrast, linear interpolation converges faster but does not guarantee convergence, and it might fail if the function is not well approximated by a linear model in the interval.
No real roots. Imaginary roots as this function does not intersect the X axis.
It has two complex roots.
There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
charles dowel
Square roots are computed using the Babylonian method, calculators, Newton's method, or the Rough estimation method. * * * * * Or the Newton-Raphson method.
x=16
It has roots x = 2.618 and x = 0.38197
No real roots
-4,3 are the roots of this equation, so for the values for which the sum of roots is 1 & product is -12
false
The roots are: x = -5 and x = -9