Question

Zeroes of the polynomial x^2 - 9 are -3 and​

Answer 1

Answer:

2x²-9x+4=0

2x²-9x+4=02x²-8x-x+4=0

2x²-9x+4=02x²-8x-x+4=02x(x-4)-1(x-4)=0

2x²-9x+4=02x²-8x-x+4=02x(x-4)-1(x-4)=0(2x-1)(x-4)=0

2x²-9x+4=02x²-8x-x+4=02x(x-4)-1(x-4)=0(2x-1)(x-4)=0x= 1/2 or x = 4.

2x²-9x+4=02x²-8x-x+4=02x(x-4)-1(x-4)=0(2x-1)(x-4)=0x= 1/2 or x = 4.Sum of zeros = 1/2+4=4 and 1/2. (-coefficient of x² /coefficient of x)

2x²-9x+4=02x²-8x-x+4=02x(x-4)-1(x-4)=0(2x-1)(x-4)=0x= 1/2 or x = 4.Sum of zeros = 1/2+4=4 and 1/2. (-coefficient of x² /coefficient of x)Product of zeros = 1/2 * 4 = 2 (constant term / coefficient of x²)

Step-by-step explanation:

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Answer 2

Answer:

0

Step-by-step explanation:

Alpha and beta = -b/a

0/1 = 0