Question

Form equation whose roots are 6+5i,3-5i

Answer 1

Step-by-step explanation:

The square root of a negative number is called an imaginary number, e.g. V-4, ... If z, = 3 + 2i and Z2=-1 +5i, express in the form a+bi, a,beR: ... Find a quadratic equation whose roots are:.

Answer 2

Answer:

The equation with given roots is $x^2-9x+(23-15i)=0$

Step-by-step explanation:

The given roots of the equation are $6+5i,3-5i$

We are required to find a quadratic equation with the given roots.

We know that,

If α and β are the roots of a quadratic equation $f(x)=0$ then the quadratic equation is expressed as

$(x-\alpha)(x-\beta)=0$

For our problem we have

$\alpha=6+5i,\beta=3-5i$

Therefore,the quadratic equation is

$[x-(6+5i)][x-(3-5i)]=0\\= > x^{2}+(6+5i)(3-5i)-[(6+5i)+(3-5i)]x=0\\= > x^2+18-5i^2-15i-9x=0\\= > x^2-9x+(23-15i)=0$

Therefore,

The equation with given roots is $x^2-9x+(23-15i)=0$

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