Question

Q3. For what value(s) of k, are the roots of quadratic equation are equal ? (k +1)x² -2(K-1 )x + 1 = 0 -

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Answer 1

Answer:

The value of k =0 or k=3

Step-by-step explanation:

Given equation   (k +1)x² -2(K-1 )x + 1 = 0

compare the equation with $ax^{2} +bx+c=0$

 ⇒ $a = (k+1)$ ,  $b= -2(k-1)$  and   $c = 1$

roots of given equation are real

        Δ = $b^{2} - 4ac$ = 0

            $[-2(k-1) } ]^{2} - 4 (k+1) (1) =0$

            4 $(k-1)^{2}$- 4$(k+1)$ =0

            $4 [ (k-1)^{2} -(k+1)] =0$$[( k-1)^{2} = k^{2} +1-2k] from (a-b)^{2} = a^{2} +b^{2} -2b]$

               $k^{2} +1- 2k -k-1=0$  

                 $k^{2} -3k =0$ ⇒   $k( k -3)=0$

                   $k= 0$  or    $(k-3)=0$

                                       $k=3$