Question

If one root of x2 - (3+2i )x +(1+3i) = 0 is
1+i then the other root is
1) 1-i 2) 2+i 3) 3+i 4) 1+3i​

Answer 1

GIVEN :

If one root of the quadratic equation is $x^2 - (3+2i )x +(1+3i) = 0$ is 1+i

TO FIND :

The other root of the given quadratic equation.

SOLUTION :

Given that one root of the quadratic equation is $x^2 - (3+2i )x +(1+3i) = 0$ is  1+i

Now we have to find the other root of the given quadratic equation.

Let $\alpha$ and $\beta$ be the two roots of the given quadratic equation.

From the given quadratic equation a=1 , b=3+2i and c=1+3i

$Sum of the roots=\alpha+\beta=-\frac{b}{a}$

Substitute the values we get,

$Sum of the roots=\alpha+\beta=-\frac{3+2i}{1}$

$=-(3+2i)$

$Sum of the roots=\alpha+\beta=-(3+2i)$

$Product of the roots=\alpha \beta=\frac{c}{a}$

Substitute the values we get,

$Product of the roots=\alpha \beta=\frac{1+3i}{1}$

$Product of the roots=\alpha \beta=1+3i$

Now from , $Sum of the roots=\alpha+\beta=-(3+2i)$

Since $\alpha=1+i$ we have that, Now sum the roots we get

$\alpha+\beta=(3+2i)$

$1+i+\beta=3+2i$

$\beta=3+2i-1-i$

$=2+i$

$\beta=2+i$

∴ other root is $\beta=2+i$

$Product of the roots=\alpha \beta=1+3i$

$=(1+i)(2+i)$

$=2+i+2i+i^2$

$=2+3i-1$

$=1+3i$

∴ $Product of the roots=\alpha \beta=1+3i$ is verified.

∴ option 2) 2+i is correct.

Answer 2

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Step-by-step explanation:

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