Question

relationship find √3x²-8x+4√3​

Answer 1

Given, quadratic polynomial is √3x2 - 8x + 4√3

= √3x2 - 6x - 2x + 4√3

= √3x2 - √3 * √3 * 2 * x - 2x + 4√3

= √3x(x - 2√3) - 2(x - 2√3)

= (x - 2√3)(√3x - 2)

The value of √3x2 - 8x + 4√3 is zero if x - 2√3 = 0 or √3x - 2 = 0

So, x = 2√3, 2/√3

Therefore, zeroes of x2 – 2x – 8 are 2√3 and 2/√3

Now, Sum of zeroes = 2√3 + 2/√3

= (2√3 * √3 + 2)/√3

= (2 * 3 + 2)/√3

= (6 + 2)/√3

= 8/√3

= -(-8)/√3

= -(Coefficient of x)/ (Coefficient of x2)

Product of zeroes = 2√3 * (2/√3)

= (2√3 * 2)/√3

= 4√3/√3

= Constant term/ (Coefficient of x2)

Hence, the relation between its zero and coefficients is verified.