Question

Solve this equation for 100 points

x² - 4x + 13 [ Don't use factor formula ]
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Answer 1

Answer:

$\displaystyle{Values \ of \ x\implies(2-3i) \ or \ (2+3i)$

Step-by-step explanation:

$\displaystyle{Given \ p(x)=x^2-4x+13}\\\\\displaystyle{Here \ we \ have \ not \ to \ use \ direct \ formula}\\\\\displaystyle{Well, \ we \ can \ spit \ 13 \ as \ 9+4}\\\\\displaystyle{Now \ p(x)=x^2-4x+9+4}\\\\\displaystyle{p(x)=x^2-4x+4+9}$

$\displaystyle{We \ can \ write \ it \ as}\\\\\displaystyle{p(x)=(x)^2-2\times2\times x+(2)^2+9}\\\\\displaystyle{Now \ using \ (x-y)^2=x^2-2xy+y^2}\\\\\displaystyle{p(x)=(x-2)^2+9}\\\\\displaystyle{We \ can \ i^2=-1 \ so \ put \ it \ here}\\\\\displaystyle{p(x)=(x-2)^2-9i^2}\\\\\displaystyle{Rewrite \ it \ as}\\\\\displaystyle{p(x)=(x-2)^2-(3i)^2}$

$\displaystyle{Now \ using \ (x^2-y^2)=(x+y)(x-y)}\\\\\displaystyle{p(x)=(x-2+3i)(x-2-3i)}\\\\\displaystyle{Now \ values \ of \ x:}\\\\\displaystyle{x-2+3i=0}\\\\\displaystyle{x=2-3i \ OR}\\\\\displaystyle{x-2-3i=0}\\\\\displaystyle{x=2+3i}\\\\\displaystyle{Thus \ we \ get \ zeroes \ of \ p(x) \ are :}\\\\\displaystyle{(2-3i) \ and \ (2+3i)$

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.