+0

# how to solve using vietas formulas?

0
78
1

Two of the roots of the equation ax^3+bx^2+cx+d=0 are  -5 and 6. Find an expression for the third root in terms of only a and c (that is, your expression should not involve b or d).

Mar 16, 2023

#1
0

Since -5 and 6 are roots of the equation, we know that (x+5) and (x-6) are factors of the cubic polynomial. This can be seen by using the factor theorem:

f(-5) = a(-5)^3 + b(-5)^2 + c(-5) + d = 0 f(6) = a(6)^3 + b(6)^2 + c(6) + d = 0

Since both expressions are equal to zero, we can use them to write:

a(-5)^3 + b(-5)^2 + c(-5) + d = a(6)^3 + b(6)^2 + c(6) + d

Simplifying this equation, we get:

-125a + 25b - 5c = 216a + 36b + 6c

Collecting like terms and simplifying further, we get:

341a + 61b = 11c

Now, using the fact that (x+5) and (x-6) are factors of the cubic polynomial, we can write:

f(x) = a(x+5)(x-6)(x-r)

where r is the third root of the polynomial.

Expanding the right-hand side of this expression, we get:

f(x) = a(x^2-x-30)(x-r)

Using the fact that the sum of the roots of a cubic polynomial is given by -b/a, we have:

-5 + 6 + r = -b/a

Simplifying this equation, we get:

r = -1 + b/a

Now, we can use the equation we derived earlier, 341a + 61b = 11c, to eliminate b and express r in terms of a and c:

r = -1 + b/a = -1 + (11c - 341a)/(61a) = (-342a + 11c)/(61a)

Therefore, the expression for the third root in terms of only a and c is:

r = (-342a + 11c)/(61a)

Mar 16, 2023