How do you solve $x + 9 = 2(x - 1)^2$ ?

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Answer 1 · 2018-04-15T19:15:23

$x_1=7/2$
$x_2=-1$

$x+9=2(x^2-2x+1)$

$x+9=2x^2-4x+2$

$-2x^2+x+4x+9-2=0$

$-2x^2+5x+7=0 //*(-1)$

$2x^2-5x-7=0$

Now use the formula $x_(1//2)=(-b+-sqrt(b^2-4ac))/(2a)$
$a=2$
$b=-5$
$c=-7$

$x_(1//2)=(5+-sqrt((25-4*2(-7))))/(2*2)$

$x_(1//2)=(5+-sqrt((25+56)))/4$

$x_(1//2)=(5+-sqrt(81))/4$

$x_(1//2)=(5+-9)/4$

$x_1=(5+9)/4=14/4=7/2$

$x_2=(5-9)/4=(-4)/4=-1$

How do you solve x + 9 = 2(x - 1)^2x + 9 = 2(x - 1)^2?