Question

If p and q are the roots of x2 – 7x – 6 = 0, find the value of p/q + q/p.
(it's x square).

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:p \: and \: q \: are \: roots \: of \: {x}^{2} - 7x - 6 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:p + q = - \dfrac{( - 7)}{1} = 7$

Also,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:pq = \dfrac{ - 6}{1} = - 6$

Now, Consider

$\rm :\longmapsto\:\dfrac{p}{q} + \dfrac{q}{p}$

$\rm \:  =  \: \dfrac{ {p}^{2} + {q}^{2} }{pq}$

$\rm \:  =  \: \dfrac{ {p}^{2} + {q}^{2} + 2pq - 2pq}{pq}$

$\rm \:  =  \: \dfrac{ {(p + q)}^{2} - 2pq}{pq}$

$\rm \:  =  \: \dfrac{ {7}^{2} - 2( - 6)}{ - 6}$

$\rm \:  =  \: \dfrac{49 + 12}{ - 6}$

$\rm \:  =  \: \dfrac{61}{ - 6}$

$\rm \:  =  \: - \: \dfrac{61}{6}$

Hence

$\rm \implies\:\boxed{\tt{ \dfrac{p}{q} + \dfrac{q}{p} = \: - \: \frac{61}{6} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac