Question

How many roots do the following equations have? -12x^2 - 25x+5 +x^3=0

Answer 1

`-12x^2 - 25x + 5 + x^(3) = 0`

`x^(3) - 12x^(2) - 25x + 5 = 0`



`x = \sqrt[3]{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) + \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3)}} + \sqrt[3]{((-(-12)^(3))/(27(1)^(3))}} + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) - \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3) - (-12)/(3(1))}}`







`x = \sqrt[3]{(114 - 2(1)/(2)) + \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} + \sqrt[3]{(114 - 2(1)/(2)) - \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} - 4`

`x = \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - 4`

`x = \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} + \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} - 4`

`x = \sqrt[3]{112(1)/(2) + √(-1751.787)} + \sqrt[3]{112(1)/(2) - √(-1751.787)} - 4`

`x = \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i} - 4`

`x = -4 + \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i}`

Answer 2

There are 3 roots of the given equation.

Explanation:

Given the equation

`-12x^2-25x+5+x^3=0`

we have to tell the number of roots of the given equation.

As the number of roots for an equation is equal to degree.

The degree of a polynomial is the highest power of its monomials with non-zero coefficients.

Hence, number of roots is the highest power in the equation.

Now, the equation is
`-12x^2-25x+5+x^3=0`

The highest power i.e degree of equation is 3.

hence, there are 3 roots of the given equation.