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I) 1,1
p(x)= k(x² - ( sum of zeroes)x + (product of zeroes)
=[x²-(1)x + (1)]
=x²-x+1
similarly
p(x)= k(x² - ( sum of zeroes)x + (product of zeroes)
=x²-4x+1
×– Quadratic polynomial –×
➜ (iv) 1 , 1 ( these are zeros )
so now,
✧ œ + ß = 1 { here we assume that œ = Alpha )
œ ß = 1
✧ œ + ß = -b = 1
a 1
✧ œ ß = c = 1
a 1
hence , b = - 1 , a = 1 , c = 1 { - b = 1 so, b = - 1 }
ax²+ bx + c = 0
x² - x + 1 = 0
➜ (vi) 4 , 1
✧ œ + ß = 4
œ ß = 1
✧ œ + ß = - b = 4
a 1
✧ œ ß = c = 1
a 1
hence, b = - 4 , c = 1 , a = 1 { - b = 4 so, b = - 4 }
ax² + bx + c = 0
x² - 4x + 1 = 0
I hope it helps you