Answered

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Solve by using square root:
4(x-1)^2+2=10

Sagot :

Konrad509idn
[tex] 4(x-1)^2+2=10 \\ 4(x-1)^2=8\\ (x-1)^2=2\\ x-1=-\sqrt2 \vee x-1=\sqrt2\\ x=1-\sqrt2 \vee x=1+\sqrt2[/tex]
Lilithidn
[tex]4(x-1)^2+2=10\\\\4(x^2-2x+1)+2-10=0\\\\4 x^2-8x+4+2-10=0\\\\4 x^2-8x-4=0\ \ / :4\\\\x^2-2x-1=0\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{2-\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2-\sqrt{ 8}}{2 }=\frac{2- \sqrt{ 4*2}}{2 }=\\\\=\frac{2-\sqrt{ 8}}{2 }=\frac{2(1- \sqrt{ 2})}{2 }=1- \sqrt{ 2}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2+\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2(1+ \sqrt{ 2})}{2 }=1+ \sqrt{ 2}[/tex]

[tex]Answer:\ x=1- \sqrt{2}\ \ or\ \ x= 1+ \sqrt{ 2}[/tex]


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