Question

plz solve this question correctly and also send it on my telegram id(send here also)
@vaishnavi25v​

Answer 1

Answer:

5)a) Given: quadratic equation kx ²

+4x+1=0, has real and distinct roots

To find the value of k

• Sol: An equation has real and distinct roots if the discriminant b²-4ac > 0

In the given equation, a=k,b=4,c=1

Therefore the discriminant is 4²

−4(k)(1)>0

16−4k>0

⟹4k<16

Therefore, k<4

c)The given equation is

x²+2(k+2)x+9k=0

For the equation to have equal roots, the discriminant must be equal to zero

⇒b²- 4ac

⇒4(k+2)²

−4(9k)=0

⇒(k+2)²

=9k

⇒k²-5k+4=0

⇒k=1,4

6)a)2x²-4x+k=0.

a = 2

b = -4

c = k

→d=b²−4ac=0

→d=(−4)²−4×2×k=0

→d=16−8k=0

→16=8k

→ 16/8=k

→k=2

7)a)kx²−5x+1=0

Here a=k,b=−5,c=1

Given that the equation has equal roots.

Therefore,

D=0

b²−4ac=0

(−5)−4(k)(1)=0

25−4k=0

⇒4k

=25

⇒k=25/4

hope this helps!!

thanks.

Answer 2

Answer:

k=25

Step-by-step explanation:

this is your answer