obojeneqk

2022-09-12

Solve the equation over the complex numbers.

$$5{x}^{2}-6x+5=0$$

$$5{x}^{2}-6x+5=0$$

empatiji2v

Beginner2022-09-13Added 18 answers

$$5{x}^{2}-6x+5=0$$

We will use the method of completing the square to find the values of x.

$$5{x}^{2}-6x+5=0\phantom{\rule{0ex}{0ex}}{x}^{2}-\frac{6}{5}x+1=0\phantom{\rule{0ex}{0ex}}{x}^{2}-2\times \frac{3}{5}+(\frac{3}{5}{)}^{2}-(\frac{3}{5}{)}^{2}+1=0\phantom{\rule{0ex}{0ex}}(x-\frac{3}{5}{)}^{2}-\frac{9}{25}+1=0\phantom{\rule{0ex}{0ex}}(x-\frac{3}{5}{)}^{2}=\frac{9}{25}-1\phantom{\rule{0ex}{0ex}}x-\frac{3}{5}=\pm \sqrt{-\frac{16}{25}}=\pm \sqrt{\frac{16}{25}{i}^{2}}\phantom{\rule{0ex}{0ex}}x-\frac{3}{5}=\pm \frac{4}{5}i$$

Therefore $$x=\frac{3}{5}+\frac{4}{5}i$$

We will use the method of completing the square to find the values of x.

$$5{x}^{2}-6x+5=0\phantom{\rule{0ex}{0ex}}{x}^{2}-\frac{6}{5}x+1=0\phantom{\rule{0ex}{0ex}}{x}^{2}-2\times \frac{3}{5}+(\frac{3}{5}{)}^{2}-(\frac{3}{5}{)}^{2}+1=0\phantom{\rule{0ex}{0ex}}(x-\frac{3}{5}{)}^{2}-\frac{9}{25}+1=0\phantom{\rule{0ex}{0ex}}(x-\frac{3}{5}{)}^{2}=\frac{9}{25}-1\phantom{\rule{0ex}{0ex}}x-\frac{3}{5}=\pm \sqrt{-\frac{16}{25}}=\pm \sqrt{\frac{16}{25}{i}^{2}}\phantom{\rule{0ex}{0ex}}x-\frac{3}{5}=\pm \frac{4}{5}i$$

Therefore $$x=\frac{3}{5}+\frac{4}{5}i$$

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