Find the values of k for which the given quadratic equation has real and distinct roots:
(i) KX²+2X+1=0
(ii) KX²+6X+1=0
(iii)X²-KX+9=0
Answers
(i) The value of K in the quadratic equation KX² + 2X + 1 = 0 is K < 1.
(ii) The value of K in the quadratic equation KX² + 6X + 1 = 0 is K < 9.
(iii) The value of K in the quadratic equation X² - KX + 9 = 0 is K > 6.
• A quadratic equation ax² + bx + c = 0 has real and distinct roots if its discriminant > 0.
• Discriminant = b² - 4ac
∴ For real and distinct roots,
b² - 4ac > 0
where a is the coefficient of x²,
b is the coefficient of x,
and c is the constant.
(i) Given quadratic equation is :
KX² + 2X + 1 = 0
Here, a = K, b = 2, c = 1
• According to the question,
b² - 4ac > 0
=> 2² - 4.K.1 > 0
=> 4 - 4K > 0
=> 4 (1 - K) > 0
=> (1 - K) > 0 / 4
=> 1 - K > 0
=> 1 > K
=> K < 1
(ii) The given quadratic equation is : KX² + 6X + 1 = 0
∴ a = K, b = 6, c = 1
• According to the question,
b² - 4ac > 0
=> 6² - 4.K.1 > 0
=> 36 - 4K > 0
=> 36 > 4K
=> 4K < 36
=> K < 36 / 4
=> K < 9
(iii) Given that, x² - KX + 9=0
∴ a = 1, b = - K, c = 9
• According to the question,
b² - 4ac > 0
=> (-K)² - 4.1.9 > 0
=> K² - 36 > 0
=> K² > 36
=> K > √36
=> K > - 6, + 6
=> K > - 6, K > 6
Therefore, the common value for K is K > 6.
[ Since 6 > - 6 ]
Answer:
ii) kx²-6x+1=0
a=k , b= -6 , c=1
b²-4ac
(-6)²-4 (k) (1)
(-36) - 4k > 0
36-4k
k= 36/4
k= 9
value of k is 9