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If the zeroes of quatradic polynomial x^2 ( a+1)x+ b are 2 and -3 then |
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Answer»
p(x) = x²+ (a + 1)x + b
when x = 2 ➝ 2²+ (a+1)2 + b ➝ 4 + 2a + 2 + b ➝ 6 +2a + b = 0 ➝ 2a + b = -6 ..........(i) when x = -3 ➝ -3² + (a+1)×(-3) + b = 0 ➝ 9 - 3a -3 + b = 0 ➝ 6 - 3a +b =0 ➝ -3a + b = -6 ...........(II) From equation 1ST and 2nd, 2a + b = -6 - 3a + b = -6 +⠀⠀--⠀⠀+ a ⠀⠀= 0 a ⠀⠀=⠀0 so, a = 0 putting value of a in equation 1st, ➝ 0+ b = -6 ➝ b = -6 ➝ b = -6 hence a = 0 and b = -6 ━━━━━━━━━━━━━━━━━━━━━━━━━ |
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