How do you solve 10-2(x-1)^2=4?

Mar 31, 2018

$x = 1 \pm \sqrt{3}$

Explanation:

This is in 'almost' completing the square format.

Subtract 10 from both sides

$- 2 {\left(x - 1\right)}^{2} = - 6$

Lets make the LHS positive: Multiply both sides by (-1)

$+ 2 {\left(x - 1\right)}^{2} = + 6$

Divide both sides by 2

${\left(x - 1\right)}^{2} = 3$

Square root both sides

$x - 1 = \pm \sqrt{3}$

$x = 1 \pm \sqrt{3}$

If you change this to decimals you will introduce rounding errors so leave it as it is.

Mar 31, 2018

$x = 1 \pm \sqrt{3}$

Explanation:

Move 10 to the other side and divide by 2
$3 = {\left(x - 1\right)}^{2}$
Take the square root.

Mar 31, 2018

$x = 1 \pm \sqrt{3}$

Explanation:

$\text{isolate } {\left(x - 1\right)}^{2}$

$\text{subtract 10 to both sides}$

$\cancel{10} \cancel{- 10} - 2 {\left(x - 1\right)}^{2} = 4 - 10$

$\Rightarrow - 2 {\left(x - 1\right)}^{2} = - 6$

$\text{divide both sides by } - 2$

$\frac{\cancel{- 2}}{\cancel{- 2}} {\left(x - 1\right)}^{2} = \frac{- 6}{- 2}$

$\Rightarrow {\left(x - 1\right)}^{2} = 3$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 1\right)}^{2}} = \pm \sqrt{3} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x - 1 = \pm \sqrt{3}$

$\text{add 1 to both sides}$

$\Rightarrow x = 1 \pm \sqrt{3} \leftarrow \textcolor{red}{\text{exact solutions}}$