Question

find tha value of k in the equation x^2-2x(1+3k)+7(3+2k)=0 if their roots are real and equal​

Answer 1

Step-by-step explanation:

For equal roots D=0

x

2

−2x(1+3k)+7(3+2k)=0

D=b

2

−4ac

={−2(1+3k)}

2

−4(1)7(3+2k)

=4(9k

2

+6k+1)−2θ(3+2k)

9k

2

−θk−20=0

(k−2)(9k+10)=0

k=2, −10/9.

Answer 2

Answer:

2

Step-by-step explanation:

Roots are real and equal when the Discriminant is zero

so, (-2(1+3k))^2 - 4*1*(7(3+2k)) = 0

4(1+9k^2 + 6k) - 4*(21+14k) = 0

4+36k^2+24k - 84 - 56k = 0

36k^2 - 32k - 80 = 0

9k^2 - 8k - 20 = 0

k will come out to be 2