Question

formula de bhashara x2-16+64=0 me ajudem

Answer 1

$x^{2}-16+64=0,~onde: \\ \\ a=1,~b=-16,~c=64 \\ \\ \Delta~=b^{2}-4.a.c \\ \Delta~=(-16)^{2}-4.1.64 \\ \Delta~=256-256 \\ \Delta~=0$

Quando delta for igual à zero, teremos duas raízes iguais. Vamos ver:

$x= \frac{-b\pm \sqrt{\Delta~}}{2.a} \\ x= \frac{16\pm \sqrt{0}}{2.1} \\ x= \frac{16\pm0}{2} \\ \\ x'= \frac{16+0}{2}= \frac{16}{2}=8 \\ x''= \frac{16-0}{2}= \frac{16}{2}=8$

As raízes são iguais à 8.

Answer 2

$x^{2}-16+64=0,~onde: \\ \\ a=1,~b=-16,~c=64 \\ \\ \Delta~=b^{2}-4.a.c \\ \Delta~=(-16)^{2}-4.1.64 \\ \Delta~=256-256 \\ \Delta~=0$

Quando delta for igual à zero, teremos duas raízes iguais. Vamos ver:

$x= \frac{-b\pm \sqrt{\Delta~}}{2.a} \\ x= \frac{16\pm \sqrt{0}}{2.1} \\ x= \frac{16\pm0}{2} \\ \\ x'= \frac{16+0}{2}= \frac{16}{2}=8 \\ x''= \frac{16-0}{2}= \frac{16}{2}=8$

As raízes são iguais à 8.