Question

1f the zeroes of the quadratic polynomial x^2 + (a + 1) x + b are 2 and -3, then.. ​

Answer 1

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1f the zeroes of the quadratic polynomial x^2 + (a + 1) x + b are 2 and -3, then.. find a and b

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GIVEN

P(X)= X²+(a+1)X+b

ZEROES OF EQUATION ARE 2, (-3)

SUM OF ZEROES=2+(-3)=(-1)

AS WE KNOW THE SUM OF ZEROES IS

$= \dfrac{ - </strong><strong>COEEFICIENT\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>OF\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong> X}{</strong><strong>COEEFICIENT\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong> OF </strong><strong>\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong>{x}^{2} } \\ \\ = \frac{ - (a + 1)}{1}$

ACCORDING TO QUESTION

-(a+1)=(-1)

=>a+1=1

=>a=0

PRODUCT OF ZEROES=(-3×2)=(-6)

AS WE KNOW THAR PRODUCT OF ZEROES=

$= \dfrac{Constant}{COEEFICIENT </strong><strong>\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong>OF</strong><strong> </strong><strong>\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong> {x}^{2} } \\ \\ = \dfrac{b}{1}$

ACCORDING TO QUESTION

b=(-6)

THEREFORE, a=0, b=(-6)

Answer 2

Step-by-step explanation:

ZEROES OF EQUATION ARE 2, (-3)

SUM OF ZEROES=2+(-3)=(-1)

AS WE KNOW THE SUM OF ZEROES IS

$= \dfrac{ - COEEFICIENT\:\: OF\:\: X}{COEEFICIENT\:\: OF \:\:{x}^{2} } \\ \\ = \frac{ - (a + 1)}{1}$

ACCORDING TO QUESTION

-(a+1)=(-1)

=>a+1=1

=>a=0

PRODUCT OF ZEROES=(-3×2)=(-6)

AS WE KNOW THAR PRODUCT OF ZEROES=

$= \dfrac{Constant}{COEEFICIENT \:\:OF \:\: {x}^{2} } \\ \\ = \dfrac{b}{1}$

ACCORDING TO QUESTION

b=(-6)

THEREFORE, a=0, b=(-6)