Question

Encontre as raízes das funções quadráticas!
$(x+5)^{2} =25$

Answer 1

Lembrando:

$(a+b)^{2}=a^{2}+2ab+b^{2}$
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$(x+5)^{2}=25\\x^{2}+2\cdot x\cdot5+5^{2}=25\\x^{2}+10x+25=25\\x^{2}+10x+25-25=0\\x^{2}+10x+0=0\\\\(a=1~~~b=10~~~c=0)$

Calculando Δ e as raízes:

$\Delta=b^{2}-4ac\\\Delta=10^{2}-4\cdot1\cdot0\\\Delta=100$

Fórmula de Bhaskara:

$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-10\pm\sqrt{100}}{2\cdot1}=\dfrac{-10\pm10}{2}\\\\\\x'=\dfrac{-10+10}{2}=\dfrac{0}{2}~~~\therefore~~~\boxed{\boxed{x'=0}}\\\\\\x''=\dfrac{-10-10}{2}=\dfrac{-20}{2}~~~\therefore~~~\boxed{\boxed{x''=-10}}$

Answer 2

Explicação passo-a-passo:

(x+5)2=25

x2+10x+25-25=0

x2+10x=0

x+5=√25

x+5=5

x=5-5

x=0

S=-b/a

S=-10/1

S=-10

x'+x"=-10-0

x"=-10

S={0,-10}

ok!