Question

If the roots of equation kx square + 2x+1=0 are real and equal , then find value of k​

Answer 1

Given, k

2

x+2x+1=0 ______ (1) where a=k, b=2, c=1

since the equation (1) has distinct roots,

b

2

−4ac>0

⇒2

2

−4k>0

⇒4−4k>0

⇒−4k>−4

⇒4k<4

⇒k<1

∴ whenever k < 1 the given equation (1) will have distinct roots.

Answer 2

Answer:

Equation

kx² + 2x + 1 = 0

Sum of roots = -(Coefficient of x)/(Coefficient of x²)

= (-2)/k

Product of roots = Constant term/Coefficient of x²

The roots of the equation are equal. So, let the roots be a each.

Thus we have

a + a = -2/k

=> 2a = -2/k

=> a = -1/k

Also,

a×a = 1/k

=> a² = 1/k

=> (-1/k)² = 1/k

=> 1/k² = 1/k

=> k² = k

=> k = 1

Hence, the value of k is 1