Question

Como resolvo esta equação?

$2,25 = \frac{1}{x(x-24)}$

Acredito ser uma equação de segundo grau, mas quero o passo-a-passo pra compreender de verdade como resolvo essa equação.

Answer 1

$2,25=\dfrac{225}{100}=\dfrac{225\div25}{100\div25}=\dfrac{9}{4}$
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$\dfrac{9}{4}=\dfrac{1}{x\cdot(x-24)}$

Se x ≠ 0 e x ≠ 24, podemos multiplicar em cruz:

$9x\cdot(x-24)=4\cdot1\\\\9x^{2}-216x=4\\\\9x^{2}-216x-4=0\\\\\Delta=b^{2}-4ac\\\Delta=(-216)^{2}-4\cdot9\cdot(-4)\\\Delta=46656+144\\\Delta=46800\\\Delta=400\cdot117\\\Delta=400\cdot9\cdot13\\\sqrt{\Delta}=\sqrt{400}\cdot\sqrt{9}\cdot\sqrt{13}\\\sqrt{\Delta}=60\sqrt{13}$

Então:

$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-(-216)\pm60\sqrt{13}}{2\cdot9}=\dfrac{6\cdot(36\pm10\sqrt{13})}{18}=\dfrac{36\pm10\sqrt{13}}{3}$

As raízes da equação são, portanto:

$x'=\dfrac{36+10\sqrt{13}}{3}~~~\therefore~~~\boxed{\boxed{x'=12+\dfrac{10}{3}\sqrt{13}}}\\\\\\x''=\dfrac{36-10\sqrt{13}}{3}~~~\therefore~~~\boxed{\boxed{x''=12-\dfrac{10}{3}\sqrt{13}}}$

Answer 2

2,25 = 1/(x² - 24x)
2,25 . (x² - 24x) = 1
2,25x² - 54x - 1 = 0
 9/4x² - 54x - 1 = 0


Δ = (-54)² - 4.(9/4).(-1)
Δ = 2916 + 9
Δ = 2925

x = (-(-54) + - √2925)/(2.9/4)
x = (54 + - 15√13)/(9/2)


x' = (54 + 15√13)/(9/2)
x' = (54 + 15√13) . 2/9
x' = (108 + 30√13)/9
x' = 108/9 + (30√13)/9
x' = 12 + (10√13)/3


x" = (54 - 15√13)/(9/2)
x" = (54 - 15√13) . 2/9
x" = (108 - 30√13)/9
x" = 108/9 - (30√13)/9
x" = 12 - (10√13)/3