Question

2x²-18x+36=0 formula de bhaskara! 8° ano! socorro

Answer 1

$<var>2x^{2} - 18x + 36 = 0</var>$

 

$<var>Delta = b^{2} - 4ac</var>$

 

$<var>Delta = (-18)^{2} - 4(2)(36) </var>$

 

$<var>Delta = 324 - 288</var>$

 

$<var>Delta = 36</var>$

 

$<var>\sqrt{Delta} = 6</var>$

 

$<var>x = \frac{-b +- \sqrt{Delta}}{2a}</var>$

 

$<var>x = \frac{18 +- \sqrt{36}}{2.2}</var>$

 

$<var>x = \frac{18 +- 6}{4}</var>$

 

$<var>x' = 3</var>$

 

$<var>x" = 6</var>$

 

$<var>S = 3,6</var>$

Answer 2

Observe que:

 

$\text{x}=\dfrac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}$

 

Onde, $\Delta=\text{b}^2-4\cdot\text{a}\cdot\text{c}$

 

Desta maneira, em $2\text{x}^2-18\text{x}+36=0$, temos, $\text{a}=2, \text{b}=-18$ e $\text{c}=36$.

 

Logo, podemos afirmar que:

 

$\text{x}=\dfrac{-(-18)\pm\sqrt{(-18)^2-4\cdot2\cdot36}}{2\cdot2}=\dfrac{18\pm6}{4}$

 

Desse modo, as raízes são:

 

$\text{x}'=\dfrac{18+6}{4}=6$

 

$\text{x}"=\dfrac{18-6}{4}=3$

 

$\text{S}=\{6, 3\}$