a and b are the zeroes of quadratic polynomial p(x)t'2-4t+3 find value of a'4b'2+a'3b'4
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solving,
zeroes are 3,1
now putting values
81+27=108
zeroes are 3,1
now putting values
81+27=108
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0
2 x 3 = 6
Step-by-step explanation:
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