How do you solve for x in simplest radical form: $2(x+3)^2+10=66$ ?

Question

2725 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-14T18:04:09

$2(x+3)^2+10=66$

$(x+3)^2 + 5 = 33$

$x^2+6x+9+5-33=0$

$x^2+6x-19=0$

Using the quadratic root formula: $(-b+-sqrt(b^2-4ac))/2a$

$x=(-6+-sqrt(36+76))/2$

$x= (-6+-4sqrt(7))/2$

$x=-3-2sqrt(7)$
or
$x=-3+2sqrt(7)$

Answer 2 · 2015-04-14T22:06:12

Alternative solution:

$2(x+3)^2 +10 = 66$

$2(x+3)^2 = 66 -10$ $color(white)"ss"$ added $-10$ on both sides

$2(x+3)^2 = 56$

$(x+3)^2 = 56/2$ $color(white)"ssssssss"$ multiplied $1/2$ on both sides

$(x+3)^2 = 28$

$x+3 = +- sqrt 28$ $color(white)"ssssss"$ there are 2 numbers whise square is 28

$x = -3 +- sqrt 28$ $color(white)"ss"$ added $-3$ on both sides

$x = -3 +- sqrt(4*7) = -3 +- sqrt4 sqrt7 = -3 +- 2sqrt7$

How do you solve for x in simplest radical form: 2(x+3)^2+10=662(x+3)^2+10=66?