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; <u>x ≈ -1.629, and 5.629</u>
(b) Using the quadratic formula, the real roots of the given function are; <u>x ≈ -1.62589, x ≈ 5.62859</u>
(c) Using three iterations, we have; the bracket is = <u>5.625</u>, and
=<u> 6.25</u>
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;
Checking for the approximation of x-value of the intercept, we have;
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;
<u>x ≈ -1.629, and 5.629</u>
(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;
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;
Which gives
The real roots are; <u>x ≈ -1.62859, and x ≈ 5.62859</u>
(c) The initial guesses are;
= 5, and
= 10
The first iteration is therefore;
f(5) × f(7.5) = 2.5 × (-10.25) = -25.625
The bracket is therefore; = <u>5</u>, and
= <u>7.5</u>
Second iteration:
f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375
The bracket is therefore; = <u>5</u>, and
= <u>6.25</u>
Third iteration
f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625
Therefore, the bracket is = 5.625, and
= 6.25
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