Question

Como resolver x²-14x+49=0 pela fórmula de Bhaskara?

Answer 1

Resolvendo a equação, temos:


$x^2-14x+49=0\\\\ \Delta=b^2-4\cdot a\cdot c\\ \Delta=(-14)^2-4\cdot1\cdot49\\ \Delta=196-196\\ \Delta=0\\\\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{14\pm\sqrt{0}}{2\cdot1}=\dfrac{14\pm0}{2}\Longrightarrow\begin{cases}x_1=x_2=\frac{14}{2}=7\end{cases}\\\\ S=\{7\}$

Answer 2

$\boxed{\Delta = b^2-4(a)(c)}\\\\ \Delta = (-14)^2-4(1)(49)\\\\ \Delta = 196-196\\\\ \Delta =0\\\\ X= \frac{14+ou-0}{2} \\\\ x^i = 7\\ x^i^i = 7\\\\ \boxed{S(7,7)}$