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Digiron [165]
3 years ago
14

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

Mathematics
2 answers:
WITCHER [35]3 years ago
5 0
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
Scilla [17]3 years ago
3 0
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}

Answer:\ x=1- \sqrt{2}\ \ or\ \ x= 1+ \sqrt{ 2}


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xeze [42]

Hello from MrBillDoesMath!

Answer:

No. That problem cited is one of 3 great unsolved problems of antiquity. See  https://en.wikipedia.org/wiki/Angle_trisection for details.


Regards,  

MrB

P.S.  I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!


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3 years ago
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Jlenok [28]
<h3>Answer:</h3>

\displaystyle x^{\frac{2}{3}}

<h3>Step-by-step explanation:</h3>

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6 0
3 years ago
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