Question

Find the value of k for which the equation kx^2+2x+1=0 has real and distinct roots

Answer 1

heya☺

if it has real roots

kx^2+2x+1=0

=) 'D 'discriminat =b^2-4a*c

=)2^2-4*k*1

=)4k=4

k=1.

hope it help you


@rajukumar☺

Answer 2

Answer:

The value of k is less than 1.

Step-by-step explanation:

Given : Quadratic equation, $kx^2+2x+1=0$

To find : Find the value of k for which the equation has real and distinct roots?

Solution :

Using quadratic formula,

General form - $ax^2+bx+c=0$

$D=b^2-4ac$  

Roots are real and distinct if D>0

Equation is $kx^2+2x+1=0$

where, a=k , b=2, c=1

$D=b^2-4ac>0$

$(2)^2-4(k)(1)>0$

$4-4k>0$

$4(1-k)>0$

$1>k$

or $k<1$

The value of k is less than 1.