Question

alguem pode responder essas questões com o método de completar quadrados ?? obg ;-)

Answer 1

a)

$x^{2}-12x+20=0\\x^{2}-2.x.6+20=0\\x^{2}-2.x.6+20+16-16=0\\x^{2}-2.x.6+36-16=0\\x^{2}-2.x.6+6^{2}-16=0\\(x-6)^{2}-16=0\\(x-6)^{2}=16\\x-6=\pm\sqrt{16}\\x-6=\pm4$

Teremos duas possibilidades:

$x-6=+4\\x=4+6\\x=10\\\\x-6=-4\\x=-4+6\\x=2\\\\\boxed{\boxed{S=\{2,10\}}}$

b)

$x^{2}+2x-15=0\\x^{2}+2.x.1-15=0\\x^{2}+2.x.1+1-1-15=0\\x^{2}+2.x.1+1^{2}-16=0\\(x+1)^{2}-16=0\\(x+1)^{2}=16\\x+1=\pm\sqrt{16}\\x+1=\pm4$

Raízes:

$x+1=+4\\x=4-1\\x=3\\\\x+1=-4\\x=-4-1\\x=-5\\\\\boxed{\boxed{S=\{-5,3\}}}$

c)

$x^{2}+4x=5\\x^{2}+2.x.2=5\\x^{2}+2.x.2+4-4=5\\x^{2}+2.x.2+2^{2}-4=5\\(x+2)^{2}-4=5\\(x+2)^{2}=5+4\\(x+2)^{2}=9\\x+2=\pm\sqrt{9}\\x+2=\pm3$

Raízes:

$x+2=+3\\x=3-2\\x=1\\\\x+2=-3\\x=-3-2\\x=-5\\\\\boxed{\boxed{S=\{-5,1\}}}$

d)

$2x^{2}-2x=24$

Dividindo tudo por 2:

$x^{2}-x=12\\x^{2}-2.x.(1/2)=12\\x^{2}-2.x.(1/2)+(1/4)-(1/4)=12\\x^{2}-2.x.(1/2)+(1/2)^{2}-(1/4)=12\\(x-[1/2])^{2}-(1/4)=12\\(x-[1/2])^{2}=12+(1/4)\\(x-[1/2])^{2}=(48+1)/4\\(x-[1/2])^{2}=49/4\\x-[1/2]=\pm\sqrt{49/4}\\x-(1/2)=\pm7/2$

Raízes:

$x-(1/2)=+7/2\\x=(7/2)+(1/2)\\x=8/2\\x=4\\\\x-(1/2)=-7/2\\x=(-7/2)+(1/2)\\x=-6/2\\x=-3\\\\\boxed{\boxed{S=\{-3,4\}}}$