Find the value of k for which the quadratic equation kx² 2x+1=0 has two real and distinct roots

Find the value of k for  the following quadratic equation so that it has two equal roots kx2 + 2x - 1 = 0

The given quadric equation is kx2 + 2x + 1 = 0, and roots are real and distinct

Then find the value of k.

Here,

a = k, b = 2 and c = 1

As we know that D = b2 - 4ac

Putting the value of a = k, b = 2 and c = 1

D = (2)2 - 4 x (k) x (1)

= 4 - 4k

The given equation will have real and distinct roots, if D > 0

4 - 4k > 0

Now factorizing of the above equation

4 - 4k > 0

4k < 4

k < 4/4

k < 1

Now according to question, the value of k less than 1

Therefore, the value of k < 1.

Page 2

The given quadric equation is kx2 + 6x + 1 = 0, and roots are real and distinct.

Then find the value of k.

Here,

a = k, b = 6 and c = 1

As we know that D = b2 - 4ac

Putting the value of a = k, b = 6 and c = 1

D = (6)2 - 4 x (k) x (1)

= 36 - 4k

The given equation will have real and distinct roots, if D > 0

36 - 4k > 0

Now factorizing of the above equation

36 - 4k > 0

4k < 36

k < 36/4

k < 9

Now according to question, the value of k less than 9

Therefore, the value of k < 9.

Page 3

The given quadric equation is x2 - kx + 9 = 0, and roots are real and distinct

Then find the value of k.

Here,

a = 1, b = (-k) and c = 9

As we know that D = b2 - 4ac

Putting the value of a = 1, b = (-k) and c = 9

D = (-k)2 - 4 x (1) x (9)

= k2 - 36

The given equation will have real and distinct roots, if D > 0

k2 - 36 > 0

Now factorizing of the above equation

k2 - 36 > 0

k2  > 36

`k>sqrt36=+-6`

k < -6 Or k > 6

Therefore, the value of k < -6 Or k > 6.