the roots of x^3-3x^2-16c+48=0 , given that the sum of two roots is zero is
x³ - 3 x² -16 c + 48 = 0
For the cubic polynomial
A x³ + B x² + C x + D = 0
Vieta's formulas:
x₁ + x₂ + x₃ = - B / A
x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = C / A
x₁ ∙ x₂ ∙ x₃ = - D / A
Where x₁, x₂ and x₃ are the roots of this polynomial.
In this cace:
A = 1 , B = - 3 , C = 0 , D = -16 c + 48
so:
x₁ + x₂ + x₃ = - B / A = - ( - 3 ) / 1 = 3 / 1 = 3
x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = C / A = 0 / 1 = 0
x₁ ∙ x₂ ∙ x₃ = - D / A = - (-16 c + 48 ) / 1 = ( 16 c - 48 ) / 1 = 16 c - 48
When the sum of two roots is zero then:
x₁ = - x₂
Now:
x₁ + x₂ + x₃ =
- x₂ + x₂ + x₃ = 3
x₃ = 3
x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = 0
( - x₂ ) ∙ x₂ + ( - x₂ ) ∙ 3 + x₂ ∙ 3 = 0
- x₂² - 3 x₂ + 3 x₂ = 0
- x₂² = 0 Multiply both sides by -1
x₂² = 0
x₂ = 0
x₁ = - x₂
x₁ = 0
Also by Vieta's formulas:
A x³ + B x² + C x + D = ( x - x₁ ) ( x - x₂ ) ( x - x₃ )
x³ - 3 x² -16 c + 48 = ( x - x₁ ) ( x - x₂ ) ( x - x₃ ) =
( x - 0 ) ( x - 0 ) ( x - 3 ) = x ∙ x ∙ ( x - 3 ) = x² ∙ ( x - 3 ) = x³ - 3 x²
x³ - 3 x² -16 c + 48 = x³ - 3 x² Subtract x³ - 3 x² to both sides
( x³ - 3 x² ) -16 c + 48 - ( x³ - 3 x² ) = ( x³ - 3 x² ) - ( x³ - 3 x² )
-16 c + 48 = 0 Subtract 48 to both sides
-16 c + 48 - 48 = - 48
-16 c = - 48 Divide both sides by - 16
c = - 48 / - 16
c = 3
The solutions:
c = 3
x₁ = 0
x₂ = 0
x₃ = 3
One correction.
A x³ + B x² + C x + D = A ( x - x₁ ) ( x - x₂ ) ( x - x₃ )
x^3-3x^2-16c+48
= x^2(x-3)-16(c-3)
If we let c=x, then we have
(x^2-16)(x-3)
= (x-4)(x+4)(x-3)
the roots are -4,4,3 so our cubic is
x^3-3x^2-16x+48=0
or, as above, let c=3, and we have
x^2(x-3) = 0
and the roots are 0,0,3
To find the roots of the equation x^3 - 3x^2 - 16c + 48 = 0, we are given that the sum of two of the roots is zero. Let's break down the process step by step:
1. Factorize the equation:
x^3 - 3x^2 - 16c + 48 = 0
First, try to find a potential root by using the Rational Root Theorem. Possible rational roots of the equation must be a factor of the constant term (48) divided by a factor of the leading coefficient (1). In this case, the potential rational roots are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
2. Identify the two roots whose sum is zero:
Since we are given that the sum of two roots is zero, the two roots can be written as (a, -a), where a is a non-zero number.
3. Rewrite the equation using the two roots:
We can rewrite the equation with the two roots as follows:
(x - a)(x + a)(x - b) = 0
To expand the equation and simplify:
(x^2 - a^2)(x - b) = 0
4. Compare the expanded equation with the original equation:
From the expanded equation, we can see that:
-a^2(x - b) = -16c + 48
This equation gives us a relationship between a, b, and c. By comparing the coefficients, we find that -a^2 = -16 and -a^2 \cdot -b = 48.
5. Solve for a and b:
From -a^2 = -16, we get a = ±√(16). Since a cannot be zero, a = ±4.
Substituting the value of a into -a^2 \cdot -b = 48:
-16 \cdot -b = 48
Simplifying gives b = 3.
6. Write the final equation:
We can write the equation with the two roots as follows:
(x - 4)(x + 4)(x - 3) = 0
7. Find the roots:
To find the roots, we set each factor equal to zero and solve:
x - 4 = 0 => x = 4
x + 4 = 0 => x = -4
x - 3 = 0 => x = 3
The roots of the equation x^3 - 3x^2 - 16c + 48 = 0, given that the sum of two roots is zero, are x = 4, x = -4, and x = 3.