Question

a^2+b^2+c^2+3 =2(a+b+c)​

Answer 1

Answer:

3

Step-by-step explanation:

Answer 2

Answer:

$solve\:for\:a,\:a^2+b^2+c^2+3=2\left(a+b+c\right)\quad :\quad a=1+\sqrt{-b^2+2b+2c-c^2-2},\:a=1-\sqrt{-b^2+2b+2c-c^2-2}$

Step-by-step explanation:

$a^2+b^2+c^2+3=2\left(a+b+c\right)$

$a^2+b^2+c^2+3=2a+2b+2c$

$\mathrm{Subtract\:}2b+2c\mathrm{\:from\:both\:sides}$

$a^2+b^2+c^2+3-\left(2b+2c\right)=2a+2b+2c-\left(2b+2c\right)$

$a^2+b^2+c^2+3-2b-2c=2a$

$\mathrm{Subtract\:}2a\mathrm{\:from\:both\:sides}$

$a^2+b^2+c^2+3-2b-2c-2a=2a-2a$

$a^2-2a+b^2+c^2+3-2b-2c=0$

$a_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(b^2+c^2+3-2b-2c\right)}}{2\cdot \:1}$

$a_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{-b^2+2b+2c-c^2-2}}{2\cdot \:1}$

$\mathrm{Separate\:the\:solutions}$

$a_1=\frac{-\left(-2\right)+2\sqrt{-b^2+2b+2c-c^2-2}}{2\cdot \:1},\:a_2=\frac{-\left(-2\right)-2\sqrt{-b^2+2b+2c-c^2-2}}{2\cdot \:1}$

$a=1+\sqrt{-b^2+2b+2c-c^2-2},\:a=1-\sqrt{-b^2+2b+2c-c^2-2}$