Question

Please answer the given question (given in picture). Please show the process.

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{x^3+p\,x+q=0}$

Since it is a cubic equation, so, it has at most 3 roots.

One of the roots is the reciprocal of other

$\mapsto \bf{Let\,\,\,the\,\,\,roots\,\,\,be\,\,\alpha,\,\dfrac{1}{\alpha},\,\beta}$

$\bigstar\,\,\,\bf{\alpha+\dfrac{1}{\alpha}+\beta=0}$

$\bigstar\,\,\,\bf{\alpha\cdot\dfrac{1}{\alpha}+\dfrac{1}{\alpha}\cdot\beta+\beta\cdot\alpha=p}$

$\bigstar\,\,\,\bf{\alpha\cdot\dfrac{1}{\alpha}\cdot\beta=-q}$

Now,

$\tt{\beta=-q}$

So,

$\tt{\alpha+\dfrac{1}{\alpha}=q\,\,\,\,\,\,\,\,...(i)}\\\\\tt{1-\dfrac{q}{\alpha}-q\alpha=p}$

$\tt{\implies1-q\left(\dfrac{1}{\alpha}+\alpha\right)=p}$

From (i), we get,

$\tt{\implies1-q\left(q\right)=p}$

$\tt{\implies1-q^2=p}$

$\tt{\implies1=p+q^2}$

$\tt{\implies\,p+q^2=1}$