Question

if one root of the equation x²+rx-s=0 is the square of the other. prove that r³+s²+3sr-s= 0​

Answer 1

Given:

Equation $x^{2}$ +$rx$-$s$ =0

And one root of the given equation is the square root of the other.

To Prove:

We have to prove the relation: $r^{3}$+$s^{2}$+3$sr$-$s$ = 0

Solution:

The given equation is  $x^{2}$+$rx$-$s$ =0 , it is aquadratic equation.

We have, the roots of a quadratic equation a$x^{2}$+b$x$+c= 0 are $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Let us now find the roots of the given equation,

comparing the given equation   $x^{2}$+$rx$-$s$ =0 with a$x^{2}$+b$x$+c= 0 we get                    a=1, b=r, c= -s

considering the roots of the given equations as x₁ and x₂.

Now  x₁ =  $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$  =   $\frac{-r+\sqrt{r^{2}-4(1)(-s) } }{2(1)}$    =     $\frac{-r+\sqrt{r^{2}+4s } }{2}$  

And   x₂ =  $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$  =   $\frac{-r-\sqrt{r^{2}-4(1)(-s) } }{2(1)}$    =      $\frac{-r-\sqrt{r^{2}+4s } }{2}$

It is given that one root of the equation is the square of the other i.e., x₂=x₁²

                         $\frac{-r-\sqrt{r^{2}+4s } }{2}$                =   { $\frac{-r+\sqrt{r^{2}+4s } }{2}$ }²  

         

              $\frac{-r-\sqrt{r^{2}+4s } }{2}$                           =  $\frac{1}{4}${ $r^{2}$+$r^{2}$+4$s$-2$r\sqrt{r^{2}+4s }$}

              -2$r$-2$\sqrt{r^{2}+4s }$                    =  2$r^{2}$+4s-2$r\sqrt{r^{2}+4s }$

                         -4$s$-2$r$-2$r^{2}$               =  2$\sqrt{r^{2}+4s }$ (1-r)

now squaring on both sides of the equation we get,

                               (-4$s$-2$r$-2$r^{2}$)²      =  4($r^{2} +4s$)(1-r)²

 16$s^{2}$+4$r^{2}$+4$r^{4}$+16$sr$+8$r^{3}$+16$sr^{2}$        =  (4$r^{2}$+16s)(1+$r^{2}$-2r)

       4$r^{4}$+8$r^{3}$+4$r^{2}$+16$sr^{2}$+16$s^{2}$+16$sr$  =  4$r^{2}$+4$r^{4}$-8$r^{3}$+16s+16$sr^{2}$-32sr

              16$r^{3}$+$r^{2}$(16$s$+4)+16$sr$+16$s^{2}$  = $r^{2}$(16s+4)-32sr

                   16$r^{3}$+16$sr$+32sr+16$s^{2}$    =  0

                   16($r^{3}$+3$sr$+$s^{2}$-$s$)              =  0

                        r³+s²+3sr-s = 0

  Hence proved.