Question

If 2 + i √3 is a root of the equation x2 + px + q = 0,
where p, q € R, then find the ordered pair (p, q)​

Answer 1

Answer:find the solution below

Step-by-step explanation:

Answer 2

The value of p is -4 and the value of q is 7.

Step-by-step explanation:

It is given that $2+i\sqrt{3}$ is a root of the equation $x^2+px+q=0$.

According to the complex conjugate root theorem, if a+ib is a root of a polynomial p(x), then a-ib is also a root on p(x).

Using complex conjugate root theorem $2-i\sqrt{3}$ is second root of the equation $x^2+px+q=0$.

If a equation is $ax^2+bx+c=0$, then  

Sum of roots = -b/a

Product of roots = c/a

Sum of roots:

$\dfrac{-p}{1}=2+i\sqrt{3}+2-i\sqrt{3}$

$-p=4$

$p=-4$

Product of roots:

$\dfrac{q}{1}=(2+i\sqrt{3})(2-i\sqrt{3})$

$q=2^2-i^2(\sqrt{3})^2$

$q=4-(-1)(3)$

$q=4+3$

$q=7$

Therefore, the value of p is -4 and the value of q is 7.

#Learn more

If alpha and beta are the zeros of the quadratic polynomial f(x)=x2-x-4 find the value of (1) 1/alpha +1/beta -alfabeta .

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