if root of the equation x2 + k (4x + k - 1) + 2 = 0 are real and equal, find the value of k.
Answers
Answer :-
If roots of a quadratic equation are real and equal, then the value of discriminant is equal to zero.
For the given equation :-
Either ( 12k - 8 ) is zero or ( k + 1 ) is zero :-
Given :
- x² + k(4x + k - 1) + 2
- Roots are real and equal
To Find :
The value of k.
Solution :
Using the formula for discriminant and the condition for real and equal roots we can find the answer.
Required Formula :
D = b² - 4ac
Explanation :
x² + k(4x + k - 1) + 2
⇒ x² + 4kx + (k² - k + 2)
We know that if the roots are real and equal then the discriminant is 0.
ATQ,
D = b² - 4ac
where,
- a = 1
- b = 4k
- c = k² - k + 2
Substituting the values,
⇒ D = (4k)² - [4.1.(k² - k + 2)]
⇒ D = 16k² - [4(k² - k + 2)]
⇒ D = 16k² - [4k² - 4k + 8]
⇒ D = 16k² - 4k² + 4k - 8
⇒ D = 12k² + 4k - 8
For equal roots, D = 0;
⇒ 12k² + 4k - 8 = 0
Splitting the middle term,
⇒ 12k² + 12k - 8k - 8 = 0
⇒ 12k(k + 1) - 8(k + 1) = 0
⇒ (k + 1)(12k - 8) = 0
⇒ k + 1 = 0
⇒ k = -1
∴ k = -1.
⇒ 12k - 8 = 0
⇒ 12k = 8
⇒ k = 8/12
⇒ k = 2/3
∴ k = 2/3.
The value of k is -1 or 2/3.
Explore More :
- If roots are real and distinct, D > 0
- If roots are not real, D < 0
- If roots are real, D ≥ 0