Question

If the zeroes of quatradic polynomial x^2 ( a+1)x+ b are 2 and -3 then
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Answer 1

Step-by-step explanation:

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Answer 2

$\large{\underline{\bf{\purple{Given:-}}}}$

• ✦ zeroes are 2 and -3

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

• ✦ a and b

$\huge{\underline{\bf{\red{Solution:-}}}}$

p(x) = x²+ (a + 1)x + b

• x = 2 or x = -3

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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