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
the roots of x^3-3x^2-16c+48=0 , given that the sum of two roots is zero is
3 answers
One correction.
A x³ + B x² + C x + D = A ( x - x₁ ) ( x - x₂ ) ( x - x₃ )
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
= 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