Question

Solve x2 - 8x+ 8 = 0 by completing the square,

Answer 1

Answer:

$x_{1} =4+2\sqrt{2}\\x_{2}=4-2\sqrt{2}$

Step-by-step explanation:

$x^{2}$ - 8x+ 8 = 0

we divide the coefficient of the X by half :

in this case: 8x/2 = 4 , then we do the following

to the result obtained (4) squared: 4^2=16

we sum and subtract by 16 to maintain the balance of equation:

$x^{2}$ - 8x+ 16-16+8 = 0

we have:

$(x-4)^{2}$ -16 +8=0

$(x-4)^{2}$ =16-8

$(x-4)^{2}$ = 8

we write the square root on both sides of the equation:

$\sqrt{(x-4)^{2}} = \sqrt{8}$

we know:

$\sqrt{a^{2}} = abs(a)$

so we have:

abs(x-4)=$\sqrt{2^{2}2 }$

abs(x-4)=2$\sqrt{2}$

we have:

$x_{1} -4 = 2\sqrt{2} \\\\x_{2} -4 =- 2\sqrt{2}$

finally we have:

$x_{1} = 4+2\sqrt{2} \\\\x_{2} =4 - 2\sqrt{2}$