To calculate: Polinomial equation with real coefficients that has roots 1,\ 2-3i.

CheemnCatelvew

CheemnCatelvew

Answered question

2021-07-29

To calculate:
Polinomial equation with real coefficients that has roots 1, 23i.

Answer & Explanation

Nathaniel Kramer

Nathaniel Kramer

Skilled2021-07-30Added 78 answers

Formula Used:
(a+b)(ab)=a2b2
Calculation:
If the polynonial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 23i must have another root as 2+3i
Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as,
(x1)[x(23i)][x(2+3i)]=0
(x1)(x2+3i)(x23i)=0
Further use arithmetic rule.
(a+b)(ab)=a2b2
Here, a=x2, b=3i
Now, the polynomial equation is,
(x1)[(x2)2(3i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2
And i2=1
Now, the polynomial equation is,
(x1)(x24x+4+9)=0
(x1)(x24x+13)=0
(x3x24x2+4x+13x13)=0
x35x2+17x13=0
Hence, the polynomial equation of given roots 1, 23i is x35x2+17x13=0
Jeffrey Jordon

Jeffrey Jordon

Expert2022-07-06Added 2605 answers

Answer is given below (on video)

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