Question

# Find the coordinates that was of the equation $$x ^ { 3 } - p x ^ { 2 } + q x - r = 0$$, may be in A.P. and hence solve the equation $$x ^ { 3 } - 12 x ^ { 2 } + 34 x - 28 = 0$$.

Solution

## $$\text{We}\ \text{have,}$$$$\text{The}\ \text{equation}$$$${{x}^{3}}-p{{x}^{2}}-qx-r=0\ \ ......\ \ \text{in}\ \text{an}\ \text{A}\text{.P}\text{.}$$$$\text{let}\ \text{the}\ \text{roots}\ \text{of}\ a-d,\ a,\ a+d$$$$\text{Sum}\ \text{of}\ \text{roots}\ \text{=}\ \text{-}\ \dfrac{\text{coeff}\text{.of}\ {{x}^{2}}}{\text{coeff}\text{of}\ {{x}^{3}}}$$$$a-d+a+a+d=-\dfrac{-p}{1}$$$$a=p$$$$a=\dfrac{p}{3}...........(1)$$$$\text{Both}\ \text{sum}\ \text{of}\ \text{roots}\ =\ \dfrac{\text{coeff}\text{.of}\ x}{\text{coeff}\text{of}\ {{x}^{3}}}$$$$(a-d)a+a(a+d)+(a+d)(a-d)=\dfrac{q}{1}$$$${{a}^{2}}-ad+{{a}^{2}}+ad+{{a}^{2}}-{{d}^{2}}-q$$$$\Rightarrow 3{{a}^{2}}-{{d}^{2}}=q$$$$\Rightarrow 3{{\left( \dfrac{p}{3} \right)}^{2}}-{{d}^{2}}=q$$$$\Rightarrow {{p}^{2}}-3{{d}^{2}}=3q$$$$\Rightarrow 3{{d}^{2}}={{p}^{2}}-3q$$$$\Rightarrow {{d}^{2}}=\dfrac{{{p}^{2}}-3q}{3}$$$$d=\sqrt{\dfrac{{{p}^{2}}-3q}{3}}$$$$\text{now,}$$$$\text{the roots are}\text{.}$$$$a-d,\ \ a,\ \ a+d$$$$\dfrac{p}{3}-\sqrt{\dfrac{{{p}^{2}}-3q}{3}},\ \dfrac{p}{3},\ \dfrac{p}{3}+\sqrt{{{p}^{2}}-3q3}$$$$\text{Hence,}\ \text{this}\ \text{is}\ \text{the}\ \text{answer}\text{.}$$Maths

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