if one of the zeros of the quadratic polynomial (k-1)x2 + kx + 1 is -3 then the value of k is
Answers
Step-by-step explanation:
Open the above attachment.
Hope it will help you.
Mark as brilliant please.
Answer:
equation is
equation is f(X) = (K-1)x2+kx+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3)
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1 0= -9k+7.
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1 0= -9k+7.
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1 0= -9k+7. 9k = 7
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1 0= -9k+7. 9k = 7
equation is f(X) = (K-1)x2+kx+1 and one zero is (-3) in place of X in f(X) put (-3) because it is a zero after putting f(-3)= (K-1)(-3)2+k(-3)+1 0= (k-1)(-6)+k(-3)+1 0 = -6k+6-3k+1 0= -9k+7. 9k = 7 k= 7/9
Please mark it as brainlist