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Answer 1

Question :-

$\sf{If \: the \: zeros \: of \: the \: quadratic \: } \\ \sf{polynomial \: {x}^{2} + (a + 1)x + b \: are} \\ \sf{ 2 \: and \: - 3 \: then \: find \: a \: and \: b \: .}$

Answer :-

→ a = 0 and b = -6

To Find :-

→ value of a and b .

Applied condition :-

→ To find any value in any quadratic equation [ P (x) ] hence, put the given roots at the place of "x"

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Step - by - step explanation :-

Let,

$\sf{P(x) = {x}^{2} + (a + 1)x + b}$

If 2 is a root if this equation then

P(2) = 0 ,

$\small \sf{\rightarrow P(2) = {(2)}^{2} + (a + 1)2 + b = 0 } \\ \\ \rightarrow \sf{ 4 + 2a + 2 + b \: = 0} \\ \\ \rightarrow \sf{ 2a \: + b \: = - 6 \: \: \: \: \: .......(1)}$

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And ,if -3 is a roots of this equation then P(-3) = 0 ,

$\small\rightarrow \sf{ P( - 3) = {( - 3)}^{2} + (a + 1)( -3) + b \: = 0} \\ \\ \rightarrow \sf{ 9 - 3a - 3 + b \: = 0} \\ \\ \rightarrow \sf{ - 3a + b \: = - 6} \\ \\ or \\ \\ \rightarrow \sf{ 3a \: - b \: = 6} \: \: \: \: \: \: .......(2)$

Adding (1) and (2) ,

$\rightarrow \sf{2a + b + 3a - b = 6 - 6} \\ \\ \rightarrow \sf{5a \: = 0} \\ \\ \rightarrow \boxed{ \sf{ a \: = 0}}$

Now put( a = 0 ) in (2) ,

$\rightarrow \sf{3 \times 0 - b = 6} \\ \\ \rightarrow \boxed{\sf{ b \: = - 6}}$

Therefore ,

• a = 0
• b = -6

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For verification ,

Put a = 0 and b = -6 ,

$\rightarrow \sf{ {x}^{2} + x - 6 = 0} \\ \\ \rightarrow \sf{ {x}^{2} + 3x - 2x - 6 = 0} \\ \\ \rightarrow \sf{x(x + 3) - 2(x + 3) = 0} \\ \\ \rightarrow \sf{(x + 3)(x - 2) = 0} \\ \\ \sf{x \: = 2 \: and \: - 3}$

Proved .

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Hope it helps you.