Solve the equation 4x^3 + 32x^2 + 79x + 60 =0 given that the root is equal to the sum of the other two roots.
(Hint: set roots as alpha, beta, gamma)
Using the factor theorem , I tried
x = ±1, ±2, ±4, and it worked if x = -4
so by synthetic division:
4x^3 + 32x^2 + 79x + 60 = (x-4)(4x^2 + 16x + 15)
= (x-4)(2x+5)(2x+3)
(x-4)(2x+5)(2x+3)=0
x = -4, or x = -5/2, or x = -3/2
and sure enough -5/2 + (-3/2)
= -8/2
= -4
To solve the equation 4x^3 + 32x^2 + 79x + 60 = 0, assuming the root is equal to the sum of the other two roots, let's set the roots as α, β, and γ.
Now, we need to find the sum of the roots. According to the given information, it is α + β + γ.
Using Vieta's formulas, we know that the sum of the roots of a cubic equation is given by: α + β + γ = -b/a.
In our case, the coefficient of x^2 is 32, and the coefficient of x^3 is 4. Therefore, α + β + γ = -32/4 = -8.
Since we know that the root is equal to the sum of the other two roots, we can express one in terms of the others. Let's assume α = β + γ.
Now, we have two equations:
1) α + β + γ = -8
2) α = β + γ
To solve this system of equations, we can substitute equation 2 into equation 1:
(β + γ) + β + γ = -8
2β + 2γ = -8
Divide by 2:
β + γ = -4
We have now simplified the equations to:
α + β + γ = -8
β + γ = -4
From the second equation, we can express γ in terms of β:
γ = -4 - β
Now, substitute into the first equation:
α + β + (-4 - β) = -8
α - 4 = -8
α = -8 + 4
α = -4
So, we have found α = -4.
Now, substitute the values of α and γ into the equation α = β + γ:
-4 = β + (-4 - β)
Simplify:
-4 = -4
This implies that β can take any value, as there is no specific restriction on it.
Therefore, the solutions to the cubic equation 4x^3 + 32x^2 + 79x + 60 = 0, with the given condition that the root is equal to the sum of the other two roots, are: α = -4, β = any real number, and γ = (-4 - β).
To solve this equation, let's assume that the roots are alpha, beta, and gamma.
As given in the problem, the root is equal to the sum of the other two roots. Let's represent this equation algebraically:
alpha = beta + gamma
We know that the sum of the roots of a cubic equation is equal to the negative coefficient of the x squared term divided by the coefficient of the x cubed term. In this case, the sum of the roots is:
alpha + beta + gamma = -32/4 = -8
Now, let's substitute the value of alpha in terms of beta and gamma into the sum of roots equation:
beta + gamma + beta + gamma = -8
2(beta + gamma) = -8
beta + gamma = -4
So, we have the equation:
alpha = -4
Now, we can use the sum and product of roots formula for a cubic equation to find the remaining two roots.
The product of the roots of a cubic equation is given by -c/a, where c is the constant term (60) and a is the coefficient of the x cubed term (4).
Product of roots = -60/4 = -15
And, the sum of the two remaining roots can be found by subtracting alpha from the sum of the roots:
beta + gamma = -8
beta + gamma - alpha = -8 - (-4) = -4
Now, we have the product of the roots and the sum of the remaining two roots. We can solve a quadratic equation using these values to find the remaining roots.
Let's set up a quadratic equation using the sum and product of roots:
x^2 - (beta + gamma)*x + (beta * gamma) = 0
Substituting the values we have:
x^2 + 4x - 15 = 0
We can now solve this quadratic equation using factorization, completing the square, or applying the quadratic formula.