1.

Find the values of k for which the quadratic equation \((3k+1)\text{x}^2+2(k+1)\text{x}+1=0\) as equal roots. Also, find these roots.

Answer»

For a quadratic equation, ax2 + bx + c = 0, 

D = b2 – 4ac 

If D = 0, roots are equal

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

⇒ D = 4(k + 1)2 – 4(3k + 1) = 0 

⇒ k2 + 2k + 1 – 3k – 1 = 0 

⇒ k(k – 1) = 0 

⇒ k = 0, 1 

When k = 0, 

Eq. – x2 + 2x + 1 = 0 

⇒ (x + 1)2 = 0 

⇒ x = -1 

When k = 1, 

Eq. – 4x2 + 4x + 1 = 0 

⇒ (2x + 1)2 = 0 

⇒ x = -1/2



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