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( k + 4)x^2 + (k+1)x + 1 = 0
On comparing the given equation with general equation, we get
a = (k +4)
b = (k +1)
c = 1
For equal roots,
b^2 - 4ac = 0
=> (k+ 1)^2 - 4(k + 4)(1) = 0
=> k^2 + 2k + 1 - 4k - 16 = 0
=> k^2 - 2k - 15 = 0
=> k^2 - 5k + 3k - 15 = 0
=> k(k - 5) + 3(k - 5) =0
=> (k-5) (k +3) = 0
k = 5 and - 3
______________________
For k = 5
Equation is 9x^2 + 6x + 1 = 0
=> 9x^2 + 3x + 3x + 1 = 0
=> 3x ( 3x + 1) + 1(3x + 1) = 0
=> (3x + 1) (3x + 1) = 0
Roots are - 1/3 and - 1/3
______________________
For k = - 3
Equation is x^2 - 2x + 1 = 0
=> x^2 - x - x + 1 = 0
=> x ( x - 1) - 1(x - 1) = 0
=> (x - 1)(x - 1) = 0
Roots are 1 and 1
On comparing the given equation with general equation, we get
a = (k +4)
b = (k +1)
c = 1
For equal roots,
b^2 - 4ac = 0
=> (k+ 1)^2 - 4(k + 4)(1) = 0
=> k^2 + 2k + 1 - 4k - 16 = 0
=> k^2 - 2k - 15 = 0
=> k^2 - 5k + 3k - 15 = 0
=> k(k - 5) + 3(k - 5) =0
=> (k-5) (k +3) = 0
k = 5 and - 3
______________________
For k = 5
Equation is 9x^2 + 6x + 1 = 0
=> 9x^2 + 3x + 3x + 1 = 0
=> 3x ( 3x + 1) + 1(3x + 1) = 0
=> (3x + 1) (3x + 1) = 0
Roots are - 1/3 and - 1/3
______________________
For k = - 3
Equation is x^2 - 2x + 1 = 0
=> x^2 - x - x + 1 = 0
=> x ( x - 1) - 1(x - 1) = 0
=> (x - 1)(x - 1) = 0
Roots are 1 and 1
Answered by
2
Now for equal roots D = 0
a = k + 4
b = k + 1
c = 1
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