Find the value of k for which quadratic equation has real and equal roots x2+k(4x+k-1)+2
Answers
x2 + 4Kx + K(K – 1) + 2 = 0
For real and equal roots, D = 0
⇒ D = b2 – 4ac = 0
⇒ (4K)2 – 4 [1] [ K(K – 1) + 2 ] = 0
16K2 – 4 [ K2 – K + 2 ] = 0
16K2 – 4K2 + 4K – 8 = 0
12K2 + 4K – 8 = 0
3K2 + K – 2 = 0
3K2 + 3K – 2K – 2 = 0
⇒ 3K ( K +1 ) – 2 ( K + 1) = 0
⇒ ( K +1 ) ( 3K – 2) = 0
k= -1 , 2/3
Question:
Find the value of k for which the quadratic equation x² + k( 4x + k - 1 ) + 2 = 0 has equal roots.
Answer:
k = -2 , 1
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 ;
x² + k(4x+k-1) + 2 = 0
=> x² + 4kx + k² - k + 2 = 0
Clearly , we have ;
a = 1
b = 4k
c = k² - k + 2
We know that ,
The quadratic equation will have real and equal roots if its discriminant is equal to zero .
=> D = 0
=> (4k)² - 4•1•(k²-k+2) = 0
=> 16k² - 4(k²-k+2) = 0
=> 4(2k²-k²+k-2) = 0
=> k² + k - 2 = 0
=> k² + 2k - k - 2 = 0
=> k(k+2) - (k+2) = 0
=> (k+2)(k-1) = 0
=> k = -2,1