Question

Find the zeros of the quadratic polynomial ( 8x²- 4 ) and verify the relation between zeros and coefficients. ​

Answer 1

Step-by-step explanation:

Let f(x) = 8x2 – 4 = 4 ((√2x)2 – (1)2) = 4(√2x + 1)(√2x – 1) To find the zeroes, set f(x) = 0 (√2x + 1)(√2x – 1) = 0 (√2x + 1) = 0 or (√2x – 1) = 0 x = (-1)/√2 or x = 1/√2 So, the zeroes of f(x) are (-1)/√2 and x = 1/√2 Again, Sum of zeroes = -1/√2 + 1/√2 = (-1+1)/√2 = 0 = -b/a = (-Coefficient of x)/(Cofficient of x2) Product of zeroes = -1/√2 x 1/√2 = -1/2 = -4/8 = c/a = Constant term / Coefficient of x2

Answer 2

zero of this is 1/2

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