If the zero's of x^2-kx+6 are in ratio 3:2, find the value of k
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Since the zeroes are in ratio 3:2, we can assume them to be 3a and 2a.
So, p(x)= x² - kx + 6 = x² - (sum of zeroes)x + product of zeroes.
comparing both sides, we get,
6 = product of zeroes = 3a×2a =6a²
6= 6a²
a=±1
k = sum of zeroes = 3a+2a = 5a = ±5
Hence, k = ±5.
Hope you understand the approach.
All the best!
So, p(x)= x² - kx + 6 = x² - (sum of zeroes)x + product of zeroes.
comparing both sides, we get,
6 = product of zeroes = 3a×2a =6a²
6= 6a²
a=±1
k = sum of zeroes = 3a+2a = 5a = ±5
Hence, k = ±5.
Hope you understand the approach.
All the best!
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