Finding zeros of polynomial 4x^2-8x-6:
To find the zeros of the polynomial 4x^2-8x-6, we need to solve the equation 4x^2-8x-6=0. We can do this by using the quadratic formula, which is given by:

x = (-b ± sqrt(b^2-4ac))/2a

Here, a=4, b=-8, and c=-6. Substituting these values in the formula, we get:

x = (-(-8) ± sqrt((-8)^2-4(4)(-6)))/(2(4))

x = (8 ± sqrt(64+96))/8

x = (8 ± sqrt(160))/8

x = (8 ± 4sqrt(10))/8

x = 1 ± (sqrt(10))/2

So, the zeros of the polynomial 4x^2-8x-6 are 1 + (sqrt(10))/2 and 1 - (sqrt(10))/2.

Verifying the relation between zeros of polynomials:
The relation between zeros of a polynomial is given by Vieta's formulas. For a quadratic polynomial ax^2+bx+c, the sum of its zeros is -b/a and the product of its zeros is c/a. Let's verify this for the polynomial 4x^2-8x-6.

The sum of the zeros of the polynomial 4x^2-8x-6 is (1 + (sqrt(10))/2) + (1 - (sqrt(10))/2) = 2. This is equal to -(-8)/4, which is -b/a.

The product of the zeros of the polynomial 4x^2-8x-6 is (1 + (sqrt(10))/2) * (1 - (sqrt(10))/2) = 1 - (sqrt(10))^2/4 = -3/2. This is equal to c/a.

So, we have verified that the relation between zeros of the polynomial 4x^2-8x-6 satisfies Vieta's formulas.

We can write this eq. as 2x²-4x-3by shri dharacharya-(-4)±√(-4)²-(4×2×3)÷2×2=solve self