Question 955784
 Let {{{r}}} be one root. Then the other root will be {{{1/r}}}. Consequently, their product will be {{{(r)*(1/r) = 1}}}. This is true if {{{r}}} is nonzero.

Now, let {{{f(x) = (p^2+2)x^2+7x+3p= ax^2 + bx + c}}}. => {{{a=(p^2+2)}}} ,{{{b=7}}},{{{ c=3p}}}
If {{{f(x) = 0}}}, then we are given that the roots are {{{r}}} and {{{1/r}}}.
 
since product of roots is {{{ c/a=1}}}, we have

=>{{{c/a=3p/(p^2+2)=1}}}

{{{3p=1*(p^2+2)}}}

{{{0=p^2-3p+2}}}

{{{0=p^2-2p-p+2}}}

{{{0=(p^2-2p)-(p-2)}}}
{{{0=p(p-2)-(p-2)}}}
{{{0=(p-1)(p-2)}}}


=>  {{{highlight(p=1) }}} or
=> {{{highlight(p=2) }}}



since SUM OF THE ROOTS is

{{{ r + 1/r = -b/a}}}

{{{ r + 1/r = -7/(p^2+2)}}}

Computing for {{{r}}}: 

if {{{highlight(p=1) }}}

{{{ r + 1/r = -7/(1^2+2)}}}

{{{ (r^2 + 1)/r = -7/3}}}

{{{ 3(r^2 + 1) = -7r}}}

{{{ 3r^2 + 3= -7r}}}

{{{ 3r^2 +7r+ 3= 0}}}

{{{r = (1/6) (sqrt(13)-7)}}} or  {{{r = (1/6) (-7-sqrt(13))}}}

where {{{(1/6) (sqrt(13)-7) = 1/((1/6) (-7-sqrt(13)))}}}-roots are reciprocal to each other

Thus, the actual roots are:

 {{{r[1 ]= (1/6) (sqrt(13)-7) }}} or
{{{r[2 ]= (1/6 )(-7-sqrt(13))}}}

second solution:

if {{{highlight(p=2) }}}

{{{ r + 1/r = -7/(2^2+2)}}}

{{{ (r^2 + 1)/r = -7/6}}}

{{{ 6(r^2 + 1) = -7r}}}

{{{ 6r^2 + 6= -7r}}}

{{{ 6r^2 +7r+ 6= 0}}}...solving this using quadratic formula you will get


{{{r [1]= -(1/12)*i(sqrt(95)-7i)}}} or  {{{r[2] = (1/12)*i (sqrt(95)+7i)}}}

where {{{r [1]=1/r [2]}}}=> {{{-(1/12)*i (sqrt(95)-7i)=1/((1/12)*i (sqrt(95)+7i)) }}}

check our function: if {{{highlight(p=1) }}}

{{{(p^2+2)x^2+7x+3p=0}}}
{{{(1^2+2)x^2+7x+3*1=0}}}
{{{3x^2+7x+3=0}}}

roots are: {{{x = (1/6) (sqrt(13)-7)}}} or  {{{x = (1/6) (-7-sqrt(13))}}} which are reciprocal to each other


and, if {{{highlight(p=2) }}}

{{{(p^2+2)x^2+7x+3p=0}}}
{{{(2^2+2)x^2+7x+3*2=0}}}
{{{6x^2+7x+6=0}}}

roots are: => {{{x=-(1/12)*i (sqrt(95)-7i)}}} or {{{x=((1/12)*i (sqrt(95)+7i)) }}} and they are reciprocal to each other