Question

if the zeros of the polynomial x2-12X +32 are a+b and a minus b then find the value of a and b​

Answer 1

Answer:

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Answer 2

$Given \: polynomial : x^{2} - 12x + 32$

$Compare \: this \: with \: Ax^{2} +Bx + C , \\we \:get$

$A = 1 , B = -12 , C = 32$

$( a + b ) \:and \: ( a - b ) \:are \: two \: zeroes \\: of \: given \: polynomial$

$i ) Sum \:of \:the \:zeroes = \frac{-B}{A}$

$\impliee ( a + b ) + ( a - b ) = \frac{ -(-12)}{1}$

$\implies 2a = 12$

$\implies a = \frac{12}{2}$

$\implies a = 6 /: ---(1)$

$ii ) Product \:of \:the \:zeroes = \frac{C}{A}$

$\impliee ( a + b )( a - b ) = \frac{32}{1}$

$\implies a^{2} - b^{2} = 32$

$\implies 6^{2} - b^{2} = 32 \: [ From \:(1) ]$

$\implies 36 - b^{2} = 32$

$\implies 36 - 32 = b^{2}$

$\implies 4 = b^{2}$

$\implies b = \pm 2$

Case 1:

If a = 6 , b = 2 ;

a + b = 6 + 2 = 8,

a - b = 6 - 2 = 4 ,

Case 2:

If a = 6 , b = -2

a + b = 6 - 2 = 4 ,

a - b = 6 - (-2) = 6 + 2 = 8

Therefore.,

$\red { Values \: of \: ( a , b ) } \green { = ( 6 , 2) \:Or \: ( 6, -2 ) }$

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