find non zero value of k, for which the quadratic equation kx²+1 - 2(k-x)+x²=0
Answers
Given,
kx² + 1 -2(k-1)x + x²
x²(k+1) - 2(k-1) + 1
Here,
a = (k+1)
b = -2(k-1)
c = 1
For real and equal roots,
b² - 4ac = 0
[-2(k-1) ]²- 4(k+1)(1) = 0
(4k) (k-3) = 0
k = 3 or k = 0 [ Rejected. as Iqes. need non-zero value of k ]
Hence,
Non-Zero value of k is 3
Hope it helps!
Question:
Find the non zero value of k for which the quadratic equation kx² + 1 - 2(k-x) + x² = 0 has equal roots.
Answer:
k = -1/2
Note:
• An equation of degree 2 is know as quadratic equation .
• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.
• The maximum number of roots of an equation will be equal to its degree.
• A quadratic equation has atmost two roots.
• The general form of a quadratic equation is given as , ax² + bx + c = 0 .
• The discriminant of the quadratic equation is given as , D = b² - 4ac .
• If D = 0 , then the quadratic equation would have real and equal roots .
• If D > 0 , then the quadratic equation would have real and distinct roots .
• If D < 0 , then the quadratic equation would have imaginary roots .
Solution:
The given quadratic equation is ;
kx² + 1 - 2(k-x) + x² = 0
=> kx² + 1 - 2k + 2x + x² = 0
=> (k+1)x² + 2x + 1 - 2k = 0
Clearly , we have ;
a = k + 1
b = 2
c = 1 - 2k
We know that ,
The quadratic equation will have equal roots if its discriminant is equal to zero .
=> D = 0
=> 2² - 4•(k+1)•(1-2k) = 0
=> 4 - 4•(k+1)(1-2k) = 0
=> 4•[ 1 - (k+1)•(1-2k) ] = 0
=> 1 - (k - 2k² + 1 - 2k) = 0
=> 1 - (-2k² - k + 1) = 0
=> 1 + 2k² + k - 1 = 0
=> 2k² + k = 0
=> k(2k + 1) = 0
=> k = 0 , -1/2
=> k = -1/2 (appreciate value)