Solve the equation \[4{x^3} - 24{x^2} + 23x + 18 = 0\] having given that the roots are in Arithmetic Progression.

Hint: Take the roots in AP as \[a - d,a,a + d\] where d is the common difference and a is the second term of the AP. Now also use the relation between the sum of root of a cubic polynomial and product of the roots of a cubic polynomial

Complete step by step solution:
Let us consider a cubic polynomial of the form \[a{x^3} + b{x^2} + cx + d = 0\]
Where \[\alpha ,\beta ,\chi \] be the roots of the equation then it is given that the roots of the equation are in AP
Which means i can let that \[\alpha = e - f,\beta = e,\chi = e + f\]
Where e is the second term of the AP and f is the common difference between 2 terms of an AP
Again we know that the sum of all roots of the quadratic polynomial is given by
\[\alpha + \beta + \chi = \dfrac{{ - b}}{a}\& \alpha \beta \chi = \dfrac{{ - d}}{a}\]
Now if we compare the given equation with the general equation we are getting
\[\begin{array}{l}
a = 4\\
b = - 24\\
c = 23\\
d = 18
\end{array}\]
Now
\[\begin{array}{l}
\therefore \alpha + \beta + \chi = \dfrac{{ - b}}{a}\\
 \Rightarrow (e - f) + e + (e + f) = \dfrac{{ - ( - 24)}}{4}\\
 \Rightarrow 3e = \dfrac{{24}}{4}\\
 \Rightarrow e = \dfrac{6}{3}\\
 \therefore e = 2
\end{array}\]
As we have the value of e let's try to find the value of f
\[\begin{array}{l}
\therefore \alpha \beta \chi = \dfrac{{ - d}}{a}\\
 \Rightarrow (e - f)e(e + f) = \dfrac{{ - 18}}{4}\\
 \Rightarrow e\left( {{e^2} - {f^2}} \right) = \dfrac{{ - 9}}{2}\\
 \therefore {e^3} - e{f^2} = \dfrac{{ - 9}}{2}
\end{array}\]
Putting \[e = 2\] We get,
\[\begin{array}{l}
 \Rightarrow {2^3} - 2{f^2} = \dfrac{{ - 9}}{2}\\
 \Rightarrow 8 - 2{f^2} = \dfrac{{ - 9}}{2}\\
 \Rightarrow 2{f^2} = \dfrac{9}{2} + 8\\
 \Rightarrow 2{f^2} = \dfrac{{9 + 16}}{2}\\
 \Rightarrow {f^2} = \dfrac{{25}}{4}\\
 \Rightarrow f = \sqrt {\dfrac{{25}}{4}} \\
 \therefore f = \pm \dfrac{5}{2}
\end{array}\]
Which means the roots are
For positive f
\[\begin{array}{l}
\alpha = 2 - \dfrac{5}{2},\beta = 2,\chi = 2 + \dfrac{5}{2}\\
\alpha = - 0.5,\beta = 2,\chi = 4.5
\end{array}\]
For negative f
\[\begin{array}{l}
\alpha = 2 + \dfrac{5}{2},\beta = 2,\chi = 2 - \dfrac{5}{2}\\
\alpha = 4.5,\beta = 2,\chi = - 0.5
\end{array}\]
Which means in both the cases the roots are -0.5, 2, 4.5

Note:
You can also verify that your answer is correct by putting the values of roots in the equation \[\alpha \beta + \alpha \chi + \beta \chi = \dfrac{c}{a} = \dfrac{{23}}{4}\] . In these types of problems where it's up to you to choose the terms of AP, we must choose terms like \[a - d,a,a + d\] because when we add them the common difference vanishes. Similarly for GP take terms like \[ar,a,\dfrac{a}{r}\] because when we will multiply them the common ratio r vanishes, so these small things can save a lot of time while doing long calculations.