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Question: Find the condition that the zeros of the polynomial \(f(x) = x^3 + 3px^2 +3qx + r\) may be in AP...

Find the condition that the zeros of the polynomial f(x)=x3+3px2+3qx+rf(x) = x^3 + 3px^2 +3qx + r may be in AP

Explanation

Solution

To solve this problem, one should have theoretical knowledge of cubic equations and the relationship between their coefficients and roots. We will first assume suitable variables for the three roots(a−d,a,a+d)(a - d, a, a + d). Then we will apply the formula for the relationship between the roots and the coefficients of the cubic equation. From the equations, we will eliminate the variables to find the given condition. The formulas that will be used to verify this relationship are-
α+β+γ=−baαβ+βγ+γα=caαβγ=−da\mathrm\alpha+\mathrm\beta+\mathrm\gamma=\dfrac{-\mathrm b}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{\mathrm\beta\mathrm\gamma}+\mathrm{\mathrm\gamma\mathrm\alpha}=\dfrac{\mathrm c}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta\mathrm\gamma}=\dfrac{-\mathrm d}{\mathrm a}

Complete step-by-step answer :
Let the roots be a−d,a,a+da - d, a, a + d. By comparing with the f(x)f(x), the sum of roots is −3p-3p.
a−d+a+a+d=−3pa - d + a + a + d = -3p
3a=−3p3a = -3p
a=−pa = -p

By comparing with the f(x)f(x), the sum of the product of roots is 3q3q.
a(a−d)+a(a+d)+(a−d)(a+d)=3qa(a - d) + a(a + d) + (a - d)(a + d) = 3q
−p(−p−d)−p(−p+d)+(−p−d)(−p+d)=3q-p(-p -d) -p(-p + d) + (-p - d)(-p + d) = 3q
p2+dp+p2−dp+p2−d2=3qp^2 + dp + p^2 -dp + p^2 -d^2 = 3q
d2=3p2−3qd^2 = 3p^2 - 3q

By comparing with the f(x)f(x), the product of roots is −r-r.
a(a−d)(a+d)=−ra(a - d)(a + d) = -r
−p(p2−d2)=−r-p(p^2 - d^2) = -r
−p(p2−3p2+3q)=−r-p(p^2 - 3p^2 + 3q) = -r
2p2−3pq=−r2p^2 - 3pq = -r
2p3−3pq+r=02p^3 - 3pq + r = 0
This is the required condition.

Note :In such types of questions be careful in what we assume the roots. In this question, we assumed the roots as a−d,aa-d, a and a+da+d so that d gets cancelled and we get the value of a in the first equation itself. This makes the calculation a lot easier. A common mistake here is that students neglect the negative sign in the formula and often exchange the formula in a hurry.