Question

x=3+4i prove that x^4-12x^3+70x^2-204x+225=0​

Answer 1

Answer:

Step-by-step explanation:

We have,

$x=3+4i$

$\implies\,x-3=4i$

$\implies\,\left(x-3\right)^2=\left(4i\right)^2$

$\implies\,{x}^{2}-6x+9=-16$

$\implies\,{x}^{2}-6x+9+16=0$

$\implies\,{x}^{2}-6x+25=0\,\,\,\,\,\,\,\,\,\,\,\,...(1)$

Now,

${x}^{4}-12{x}^{3}+70{x}^{2}-204{x}+225$

$={x}^{4}-6{x}^{3}+25{x}^{2}-6{x}^{3}+45{x}^{2}-204{x}+225$

$={x}^{4}-6{x}^{3}+25{x}^{2}-6{x}^{3}+36{x}^{2}-150{x}+9{x}^{2}-54{x}+225$

$={x}^{2}\left({x}^{2}-6{x}+25\right)-6x\left({x}^{2}-6{x}^{2}+25\right)+9\left({x}^{2}-6{x}+25\right)$

$=\left({x}^{2}-6x+9\right)\left({x}^{2}-6{x}+25\right)$

From (1), we get,

$=\left({x}^{2}-6x+9\right)\times0$

$=0$