Find the length x please thank you

Mathematics

1 answer:

Reil [10]3 years ago

7 0

The answer is probably 6

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Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

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