write a cubic equation whose integral coefficient roots are -5 and radical 2
To write a cubic equation with integral coefficients and roots -5 and √2, we can make use of the fact that the product of the roots of a cubic equation is equal to the constant term divided by the coefficient of the highest power of x.
Let's denote the cubic equation as ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are integers.
Since the roots are -5 and √2, we can say that the equation can be factored as:
(x + 5)(x - √2)(x - α) = 0
Here, α is the other root of the cubic equation that we need to find.
Expanding the above equation gives:
(x + 5)(x - √2)(x - α) = 0
(x^2 + 5x - √2x - 5√2)(x - α) = 0
(x^2 + (5 - √2)x - 5√2)(x - α) = 0
x^3 + (5 - √2 - 5√2)x^2 + (5α - √2α - 5√2α)x - 5√2α = 0
In order to have integral coefficients for the equation, we need to ensure that the coefficients (5 - √2 - 5√2), (5α - √2α - 5√2α), and (-5√2α) are all integers.
From the first coefficient, we can set it as (5 - √2 - 5√2) = 0 to eliminate the square roots, giving:
5 - √2 - 5√2 = 0
- √2 - 5√2 = -5
-6√2 = -5
√2 = 5/6
However, this contradicts our assumption that the root √2 is irrational, so this value is not valid.
Hence, it is not possible to write a cubic equation with integral coefficients, having roots -5 and √2.