Question

10. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then * O(a) a = -7, b = -1 O (b) a = 5, b = -1 0 (c) a = 2, b = -6 (d) a = 0, b = -6​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{provided\: that:-}}}$

➻Quadratic polynomial:

$\to \tt {x^{2}+(a+1)x+b}$

➻zeroes of the polynomial:- 2 and -3

$\to \tt {p(2)=0}$

$\to \tt {p(-3)=0}$

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$\sf \blue {\underline{\underline{To\: Determine:-}}}$

➻The values of a and b.

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➻Now let us consider,p(2)=0

➻substituting in the quadratic polynomial

$\to \tt {p(2)=x^{2}+(a+1)x+b}$

$\to \tt {(2)^{2}+(a+1)(2)+b=0}$

$\to \tt {4+2a+2+b=0}$

$\to \tt {6+2a+b=0}$ --->Equation-1

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➻Now let us consider,p(-3)=0

➻substituting in the quadratic polynomial

$\to \tt {p(-3)=x^{2}+(a+1)x+b}$

$\to \tt {(-3)^{2}+(a+1)(-3)+b=0}$

$\to \tt {9-3a-3+b=0}$

$\to \tt {6-3a+b=0}$ --->Equation-2

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➻Equating both the equations.

➻Equation-1=Equation-2

$\to \tt {6+2a+b=6-3a+b}$

$\to \tt {6-6+2a+3a+b-b=0}$

$\to \tt {\cancel{6}\cancel{-6}+2a+3a{\cancel{+b}}{\cancel{-b}}=0}$

$\to \tt {5a=0}$

$\implies \tt \green {\fbox{a=0}}$

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➻Substituting the value of 'a' in either of the equations will gives us the value of 'b'.

➻Supplanting the value of 'a' in Equation-1,

$\to \tt {6+2a+b=0}$

$\to \tt {6+2(0)+b=0}$

$\to \tt {6+0+b=0}$

$\to \tt {6+b=0}$

$\implies \tt \green{\fbox{b=-6}}$

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$\sf \pink {\underline{\underline{Thereupon,}}}$

➻"Option-D" is the right answer.

➻The values of a and b are 0 and -6 respectively.

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$\large \sf \purple {\underline{\underline{Shortcut\: method:}}}$

➻In the given quadratic equation,

$\to \tt {x^{2}+(a+1)x+b}$

➻Sum of zeroes =-(a+1)

➻Product of zeroes=b

➻Now ,sum of zeroes,

$\to \tt {2+(-3)=-(a+1)}$

$\to \tt {2-3=-a-1}$

$\to \tt {-1=-a-1}$

$\to \tt {-1+1=-a}$

$\to \tt \green{\fbox{a=0}}$

➻Product of zeroes=b

$\to \tt {2×(-3)=b}$

$\to \tt {2×-3=b}$

$\to \tt \green {\fbox{b=-6}}$

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$\large \sf \orange {\underline{\underline{Hence,}}}$

➻The values of a and b are 0 and -6 respectively.

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