If one of the zeroes of the quadratic polynomial (k-1)x²+kx +15 is -3 then find the value of k
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Answered by
3
If -3 is a root of given polynomial, then f(-3)=0 .
f(-3) = (k-1)(-3)²+k(-3)+15 = 0
=> (9(k-1)-3k+15 = 0
=> 9k-9-3k+15 = 0
=> 6k -6 = 0
=> k-1 = 0
=> k = 1 .
Therefore ,K = 1
f(-3) = (k-1)(-3)²+k(-3)+15 = 0
=> (9(k-1)-3k+15 = 0
=> 9k-9-3k+15 = 0
=> 6k -6 = 0
=> k-1 = 0
=> k = 1 .
Therefore ,K = 1
Shaik96:
Check 4 th line in the process
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Answered by
3
Heya.
Your answer is here.
First of all, zero of a polynomial is that value of x which makes whole polynomial equal to zero.
It is given that the zero of polynomial is -3. So,
(k-1)(-3)²+k(-3) +15 = 0
9 (k-1) - 3k + 15 = 0
9k - 9 - 3k + 15=0
6k + 6 = 0
6k = -6
k -1.
Hence, the value of k is -1.
Hope it helps
Your answer is here.
First of all, zero of a polynomial is that value of x which makes whole polynomial equal to zero.
It is given that the zero of polynomial is -3. So,
(k-1)(-3)²+k(-3) +15 = 0
9 (k-1) - 3k + 15 = 0
9k - 9 - 3k + 15=0
6k + 6 = 0
6k = -6
k -1.
Hence, the value of k is -1.
Hope it helps
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