Question

If one of the zeros of the quadratic polynomial (k-1)X*2+kx+2 is 3, then the value of k is

Answer 1

Answer:

k = 7/12

Step-by-step explanation:

Given : $(k-1)x^2 + kx + 2$

Comparing the above equation with the standard equation - $ax^2 + bx + c$, we get-

a= k-1

b= k

c= 2

Now, let one zero = y ; other zero = 3 ( given)

We know that , sum of zeroes = -b/c

⇒ y+3 = -k/k-1........................................(i)

Also, product of zeroes = c/a

⇒ 3y = 2/k-1

or y = 2/3(k-1)..........................................(ii)

Using (ii) in (i) , we get

2/3(k-1) +3 = -k/k-1

⇒ 2/3(k-1) + k/(k-1) = -3

⇒ 2 + 3k / 3(k-1) = -3

⇒ 2 + 3k = -9k +9

∴ k = 7/12

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Answer 2

Answer:

k = 4/3

Step-by-step explanation:

(K-1) x2+kx+1 = 0

(K-1) (-3)2-3k+1 = 0

(K-1) 9-3K+1 = 0

9K-9-3K+1=0

9K-8-3K=0

9K-3K=8

6K=8

K= 8/6

K= 4/3

Hope it helps!