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प्रश्न
Find the zeroes of the polynomial q(x) = 8x2 − 2x − 3. Hence, find a polynomial whose zeroes are 2 less than the zeroes of q(x).
योग
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उत्तर
q(x) = 8x2 − 2x − 3
= 8x2 + 4x − 6x − 3 = 0
= 4x(2x + 1) − 3(2x + 1) = 0
= (2x + 1) (4x − 3) = 0
Zeroes are x = `(-1)/2, x = 3/4`
New zeroes α = `(-1)/2 - 2 = (-5)/2`
β = `3/4 - 2 = (-5)/4`
Sum of zeroes = `(-15)/4`
Product of zeroes = `25/8`
New polynomial = k[x2 − (Sum)x + Product]
= `k(x^2 + 15/4 x + 25/8)`
= 8x2 + 30x + 25
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