Question

Find the values of k for which the following equation has real and equal roots.
\[\left( k+1 \right){{x}^{2}}-2\left( k-1 \right)x+1=0\]

Answer 1

Hint: In this question, we first need to find the discriminant of the given quadratic equation. Then equate the discriminated to zero as the roots are real and equal. Now, we get a quadratic equation in k which on simplifying further gives the value of k.

Complete step-by-step answer:
Now, the given quadratic equation in the question is
\[\left( k+1 \right){{x}^{2}}-2\left( k-1 \right)x+1=0\]
As we already know that the discriminant of a quadratic equation \[a{{x}^{2}}+bx+c=0\]
is given by the formula
\[D=\sqrt{{{b}^{2}}-4ac}\]
Now, on comparing the given quadratic equation in the question with the standard form we get,
\[\begin{align}
  & a=k+1 \\
 & b=-2\left( k-1 \right) \\
 & c=1 \\
\end{align}\]
Now, on substituting these values in the discriminant formula we get,
\[\Rightarrow D=\sqrt{{{b}^{2}}-4ac}\]
Now, on substituting the values of a, b, c we get,
\[\Rightarrow D=\sqrt{{{\left( -2\left( k-1 \right) \right)}^{2}}-4\times \left( k+1 \right)\times 1}\]
Now, this can be further written as
\[\Rightarrow D=\sqrt{4{{\left( k-1 \right)}^{2}}-4\left( k+1 \right)}\]
Now, on taking the common term out we can further write it as
\[\Rightarrow D=2\sqrt{{{\left( k-1 \right)}^{2}}-\left( k+1 \right)}\]
Now, this can be further expanded and written as
\[\Rightarrow D=2\sqrt{{{k}^{2}}+1-2k-\left( k+1 \right)}\]
Now, this can also be written as
\[\Rightarrow D=2\sqrt{{{k}^{2}}-3k}\]
Now, as given in the question that it has real and equal roots it means that the discriminant of that quadratic equation should be zero.
\[\Rightarrow D=0\]
Now, on further substitution we get,
\[\Rightarrow 2\sqrt{{{k}^{2}}-3k}=0\]
Let us now do squaring on both sides to simplify it further
\[\Rightarrow 4\left( {{k}^{2}}-3k \right)=0\]
Now, on dividing both sides with 4 we get,
\[\Rightarrow {{k}^{2}}-3k=0\]
Now, on factorising the above equation we get,
\[\Rightarrow k\left( k-3 \right)=0\]
Now, from the above equation we get,
\[\Rightarrow k=0\]
\[\Rightarrow k-3=0\]
\[\Rightarrow k=3\]
\[\therefore k=3,0\]
Hence, the values of k are 0, 3.

Note: Instead of factoring the quadratic equation obtained in k we can also find the values of k by using the direct formula and simplify further. Both the methods give the same result.
It is important to note that while calculating the discriminant and value of k we should neglect any of the terms or do any calculation mistakes because it changes the corresponding equation and so the final result.