Question

Determine the real roots of f (x) = −0.6x2 + 2.4x + 5.5:(a) Graphically.(b) Using the quadratic formula.(c) Using three iterations of the bisection method to determinethe highest root. Employ initial guesses of xl = 5 and xu = 10.Compute the estimated error εa and the true error εt after eachiteration.

Answer 1

The three methods used to find the real roots of the function are,

graphically, the quadratic formula, and by iteration.

The correct vales are;

(a) Graphically, the roots obtained are; x ≈ -1.629, and 5.629

(b) Using the quadratic formula, the real roots of the given function are; x ≈ -1.62589, x ≈ 5.62859

(c) Using three iterations, we have; the bracket is $x_l$ = 5.625, and $x_u$ = 6.25

Reasons:

The given function is presented as follows;

f(x) = -0.6·x² + 2.4·x + 5.5

(a) The graph of the function is plotted on MS Excel, with increments in the

x-values of 0.01, to obtain the approximation of the x-intercepts which are

the real roots as follows;

$\begin{array}{|c|cc|}x&&f(x)\\-1.63&&-0.00614\\-1.62&&0.03736\\5.62&&0.03736 \\5.63&&-0.00614\end{array}\right]$

Checking for the approximation of x-value of the intercept, we have;

$x = -1.63 + \dfrac{0 - (-0.00614)}{0.0376 - (-0.00614)} \times (-1.62-(-1.63)) \approx -1.629$

Therefore, based on the similarity of the values at the intercepts, the x-

values (real roots of the function) at the x-intercepts (y = 0) are;

x ≈ -1.629, and 5.629

(b) The real roots of the quadratic equation are found using the quadratic

formula as follows;

The quadratic formula for finding the roots of the quadratic equation

presented in the form f(x) = a·x² + b·x + c, is given as follows;

$x = \mathbf{ \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}}$

Comparison to the given function, f(x) = -0.6·x² + 2.4·x + 5.5, gives;

a = -0.6, b = 2.4, and c = 5.5

Therefore, we get;

$x = \dfrac{-2.4\pm \sqrt{2.4^{2}-4\times (-0.6)\times 5.5}}{2\times (-0.6)} = \dfrac{-2.4\pm\sqrt{18.96} }{-1.2} = \dfrac{12 \pm\sqrt{474} }{6}$

Which gives

The real roots are; x ≈ -1.62859, and x ≈ 5.62859

(c) The initial guesses are;

$x_l$ = 5, and $x_u$ = 10

The first iteration is therefore;

$x_r = \dfrac{5 + 10}{2} = 7.5$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{10- 5}{10 + 5} \right | \times 100\% = 33.33\%$

$True \ error, \ \epsilon _t = \left|\dfrac{5.62859 - 7.5}{5.62859} \right | \times 100\% = 33.25\%$

f(5) × f(7.5) = 2.5 × (-10.25) = -25.625

The bracket is therefore; $x_l$ = 5, and $x_u$ = 7.5

Second iteration:

$x_r = \dfrac{5 + 7.5}{2} = 6.25$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{7.5- 5}{7.5+ 5} \right | \times 100\% = 20\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 6.25}{5.62859} \right | \times 100\%} \approx 11.04\%$

f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375

The bracket is therefore; $x_l$ = 5, and $x_u$ = 6.25

Third iteration

$x_r = \dfrac{5 + 6.25}{2} = 5.625$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{5.625- 5}{5.625+ 5} \right | \times 100\% = 5.88\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 5.625}{5.62859} \right | \times 100\%} \approx 6.378 \times 10^{-2}\%$

f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625

Therefore, the bracket is $x_l$ = 5.625, and $x_u$ = 6.25

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