Question

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

Answer 1

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

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.