What are the roots of the polynomial equation 3x^4+12x^2=6x^3
1 answer:
Factor the equation to give
By the zero product property, either x^2=0 or (3x^2+12-6x)=0 or both.
If x^2=0, then x=0
if (3x^2+12-6x)=0, we use the quadratic formula to solve for x
where x=1 ± √ (3) i
Answer: the roots of <span>3x^4+12x^2=6x^3 are {0 (multiplicity 2), 1+√3 i, 1-√3 i}</span>
By the zero product property, either x^2=0 or (3x^2+12-6x)=0 or both.
If x^2=0, then x=0
if (3x^2+12-6x)=0, we use the quadratic formula to solve for x
where x=1 ± √ (3) i
Answer: the roots of <span>3x^4+12x^2=6x^3 are {0 (multiplicity 2), 1+√3 i, 1-√3 i}</span>
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The solution to the quadratic equation are option C) x= -5 and option E) x=3.
<u>Step-by-step explanation</u>:
The given quadratic equation is x²+2x-15 = 0.
Using the factorization method,
- product of the roots should be -15.
- Sum of the roots should be 2.
⇒ -15 = 5 -3
⇒ 2 = 5+(-3)
(x+5)(x-3) = 0
Therefore, x = -5 and x = 3