Question

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

Answer 1

$\text{We}\text{have,}$

$\text{The}\text{equation}$

$x^3-p{x}^2-qx-r=0......\text{in}\text{an}\text{A}\text{.P}\text{.}$

$\text{let}\text{the}\text{roots}\text{of}a-d,a,a+d$

$\text{Sum}\text{of}\text{roots}\text{=}\text{-}\frac{\text{coeff}\text{.of}{x}^2}{\text{coeff}\text{of}{x}^3}$

$a-d+a+a+d=-\frac{-p}{1}$

$a=p$

$a=\frac{p}{3}...........\left(1\right)$

$\text{Both}\text{sum}\text{of}\text{roots}=\frac{\text{coeff}\text{.of}x}{\text{coeff}\text{of}{x}^3}$

$(a-d)a+a(a+d)+(a+d)(a-d)=\frac{q}{1}$

$a^2-ad+a^2+ad+a^2-d^2-q$

$\Rightarrow 3a^2-d^2=q$

$\Rightarrow 3{\left(\frac{p}{3}\right)}^2-d^2=q$

$\Rightarrow p^2-3d^2=3q$

$\Rightarrow 3d^2=p^2-3q$

$\Rightarrow d^2=\frac{p^2-3q}{3}$

$d=\sqrt{\frac{p^2-3q}{3}}$

$\text{now,}$

$\text{the roots are}\text{.}$

$a-d,a,a+d$

$\frac{p}{3}-\sqrt{\frac{p^2-3q}{3}},\frac{p}{3},\frac{p}{3}+\sqrt{p^2-3q3}$

$\text{Hence,}\text{this}\text{is}\text{the}\text{answer}\text{.}$