Question

Find the Sum and product of the roots of the given equation.

1. 4x^2 - 3x + 9 = 0
2. 3x^3 + 5x^2 - x = 3
3. x^5 - 4x^3 = 6

Answer 1

1. The sum of the roots of the equation 4x² - 3x + 9 = 0 is
`\( (3)/(4) \)`, and the product is
`\( (9)/(4) \)`.

2. For the equation 3x³ + 5x² - x = 3, both the sum and product of the roots are complex.

3. The equation x⁵ - 4x³ = 6 has roots with a sum of 0 and a product of -6.

Step-by-step explanation:

In the first equation,
`\(4x^2 - 3x + 9 = 0\)`, we can find the sum and product of the roots using Viète's formulas. For a quadratic equation
`\(ax^2 + bx + c = 0\)`, the sum of the roots
`(\(r₁\)` and
`\(r₂\))` is given by
`\( (-b)/(a) \)` and the product is
`\( (c)/(a) )\`. Applying this to the given equation, the sum is
`\( (3)/(4) \)` and the product is
`\( (9)/(4) \)`.

Moving to the second equation
`\(3x^3 + 5x^2 - x = 3\)`, the roots can be complex. Viète's formulas still apply, but now for a cubic equation
`\(ax^3 + bx^2 + cx + d = 0\)`, the sum of the roots is
`\( (-b)/(a) \)` and the product is
`\( (d)/(a) \)`. Here, we find both the sum and product to be complex.

For the third equation
`\(x^5 - 4x^3 = 6\)`, it's interesting to note that the sum of the roots is 0, and the product is -6. Viète's formulas for a quintic equation
`\(ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0\`) yield the sum
`\( (-b)/(a) \)` and the product
`\( (-f)/(a) \)`. In this case, the sum is 0, reflecting the symmetric nature of the roots, and the product is -6.