Question

one root of the quadratic equation (k+1)x^2+kx-21=0 is -7/3 find the value of k​

Answer 1

Answer:

Given x

2

−2x(1+3k)+7(3+2k)=0 has equal roots

As we know that

For the quadratic equation to have equal roots discriminant should be zero

⟹(−2(1+3k))

2

−4(1)(7(3+2k))=0

⟹4(1+3k)

2

−28(3+2k)=0

⟹(9k

2

+6k+1)−21−14k=0

⟹9k

2

−8k−20=0

⟹9k

2

−18k+10k−20=0

⟹9k(k−2)+10(k−2)=0

⟹(9k+10)(k−2)=0

⟹k=−

9

10

,2

Step-by-step explanation:

I HOPE IT HELP MARK AS BRAINLY ✅✅

Answer 2

Step-by-step explanation:

(k+1)x²+kx-21 =0

x=-7/3

(k+1)(-7/3)²+k(-7/3)-21=0

(k+1)(49/9)+-7k/3-21=0

49k+49/9-7k/3-21=0

49k+49-21k-189=0

28k-140=0

28k=140

k=5