Find the Value of k for Which the Quadratic Equation Has Equal Roots and Also Find the Roots

Question:

\( (3k+1)x^2 + 2(k+1)x + 1 = 0 \)

Find the value of \(k\) for which the equation has equal roots. Also, find the roots.

Solution

For equal roots, the discriminant must be zero.

\( D=b^2-4ac=0 \)

Here,

\( a=3k+1,\quad b=2(k+1),\quad c=1 \)

Therefore,

\( [2(k+1)]^2-4(3k+1)(1)=0 \)

\( 4(k+1)^2-4(3k+1)=0 \)

\( (k+1)^2-(3k+1)=0 \)

\( k^2+2k+1-3k-1=0 \)

\( k^2-k=0 \)

\( k(k-1)=0 \)

\( k=0 \quad \text{or} \quad k=1 \)

Finding the Equal Roots

For equal roots,

\( x=-\frac{b}{2a} \)

When \(k=0\):

\( x=-\frac{2}{2}=-1 \)

Equal roots: \( -1,\,-1 \)

When \(k=1\):

\( x=-\frac{4}{2(4)}=-\frac{1}{2} \)

Equal roots: \( -\frac{1}{2},\,-\frac{1}{2} \)

Answer

The equation has equal roots for

\( \boxed{k=0 \text{ or } k=1} \)

Corresponding equal roots are:

• \( k=0 \Rightarrow x=-1 \)
• \( k=1 \Rightarrow x=-\frac{1}{2} \)