Question

find the value of k for which the quadratic equation (3k+1)x^2+2(k+1)x+1=0 has equal roots also find its root​

Answer 1

Equation-

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

Since the roots are equal discriminant (d) is equal to 0

Discriminant-

$\boxed{ \rm \: d = {b}^{2} - 4ac}$

Here,

• a=3k+1
• b=2(k+1)
• c=1
• d=0

$\rm 0 = {(2(k + 1))}^{2} - 4 \times (3k + 1 )\times 1 \\ \rm \implies \: 0 = 4( {k}^{2} + 1 + 2k - 3k - 1) \\ \rm \implies {k}^{2} - k = 0 \\ \implies \rm k(k - 1) = 0 \\ \implies \boxed{ \rm \: k = 0 \: or \: 1}$

Now we have to find the roots of the equation

Then,

Using factorization

If we take k=0

Then,

• a=3*0+1=1
• b=2(0+1)=2
• c=1
• d=0

$\rm {x}^{2} + 2x + 1 = 0 \\ \implies \rm {x}^{2} + x + x + 1 = 0 \\ \implies \rm \: x(x + 1) + 1(x + 1) = 0 \\ \implies \boxed{ \rm \: x = - 1}$

If we take k=1

Then,

• a=3*1+1=3
• b=2*2=4
• c=1

$\rm3 {x}^{2} + 4x + 1 = 0 \\ \rm \implies 3 {x}^{2} + 3x + x + 1 = 0 \\ \implies \rm 3x(x + 1) + 1(x + 1) \\ \implies \boxed{ \rm \: x = \frac{1}{3}or - 1 }$