Question

35| Determine as raízes das funções quadráticas,
a) y= x²-16
b) y= -x²+4x
c) y= x²-13x + 40
d) y = 4 x²- 8x + 5​

Answer 1

Explicação passo-a-passo:

a) $y=x^2-16$

$x^2-16=0$

$x^2=16$

$x=\pm\sqrt{16}$

$x'=4$

$x"=-4$

b) $y=-x^2+4x$

$-x^2+4x=0$

$x\cdot(-x+4)=0$

$x'=0$

$-x+4=0~\longrightarrow~x"=4$

c) $y=x^2-13x+40$

$x^2-13x+40=0$

$\Delta=13^2-4\cdot1\cdot40$

$\Delta=169-160$

$\Delta=9$

$x=\dfrac{-(-13)\pm\sqrt{9}}{2\cdot1}=\dfrac{13\pm3}{2}$

$x'=\dfrac{13+3}{2}~\longrightarrow~x'=\dfrac{16}{2}~\longrightarrow~x'=8$

$x"=\dfrac{13-3}{2}~\longrightarrow~x"=\dfrac{10}{2}~\longrightarrow~x"=5$

d) $y=4x^2-8x+5$

$4x^2-8x+5=0$

$\Delta=(-8)^2-4\cdot4\cdot5$

$\Delta=64-80$

$\Delta=-16$

Não há raízes reais

Answer 2

Resposta:

Explicação passo-a-passo:

a) y= x²-16

x²=16

$\boxed{\boxed{x=+ \:ou\: -4}}$

b) y= -x²+4x

-x²+4x =0

x(-x+4)=0

$\boxed{\boxed{x=0}}$

-x+4=0

-x=-4

$\boxed{\boxed{x=4}}$

c) y= x²-13x + 40

 (x-5)(x-8)

x-5=0

$\boxed{\boxed{x=5}}$

x-8=0

$\boxed{\boxed{x=8}}$

d) y = 4 x²- 8x + 5​

$\Delta=b^{2} -4ac\\\Delta=(-8)^{2} -4(4)(5)\\\Delta=64-80\\\Delta=-16\\\boxed{\boxed{ =>>esta \: func\~ao\: n\~ao \:possui\: rai\´zes \: reais<<=}}$