Question

if 3+i is root of x²-ax+b=0then find a and b​

Answer 1

Answer:

In a quadratic equation with real co-efficients, complex roots occurs in pairs (root and its conjugate).

For real a,b, given 3+i is a root of the equation x2+ax+b=0......(1), then 3−i also be root of the same equation.

Now from the relation between the roots and co-efficients of (1) we get,

(3+i)+(3−i)=−a

or, a=−6.

Answer 2

Answer:

a=6

b=10

Step-by-step explanation:

Since we know that complex roots are in pair of conjugate.

so if one roots is 3+i ,the other root will be 3-i

so two roots of this eqn are 3+i and 3-i

$sum \: of \: roots =a \: \: \: \: so \\ 3 + i + 3 - i = 6$

therfore a=6.

$product \: of \: roots = b \: \: \: so \\ (3 + i)(3 - i) = 9 - 3i + 3i - {i}^{2} \\ 9 - ( - 1) = 10$

therfore b=10

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