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.