How do you solve using the completing the square method $0= x^2 + 10x + 5$ ?

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Answer 1 · 2016-05-04T18:36:44

$color(blue)(x~~ -9.472" and "-0.528" to 3 decimal places")$

Let k be the constant of correction

Write as $y=x^2+10x+5+k$ At this stage k=0

$y=(x^(color(red)(2))+10x)+5+k$

Take the square from $x^(color(red)(2)$ and move it outside the bracket

$y=(x+10x)^(color(red)(2))+5+k$

remove the $x$ from $10x$

$y=(x+10)^2+5+k$

halve the 10

$y=(x+ color(magenta)(5))^2+5+k$
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider $y=ax^2+bx+c$ written in the form $y=a(x+color(magenta)(b/(2a)))^2+c+k$

From within the bracket we have $k= (-1)xx a xx(color(magenta)(b/(2a)))^2$

The $axx$ is from the $a$ outside the bracket but in your case $a=1$

so $k=(-1)xx(color(magenta)(+5))^2 = -25$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(brown)(y=(x+5)^2+5+k)" "color(green)(->" "y=(x+5)^2+5-25$

$y=(x+5)^2-20$
'~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("solving for "x" when "y=0)$

$=>0=(x+5)^2-20$

$sqrt(20)=sqrt((x+5)^2)$

$=>x+5=+-2sqrt(5)$

$=>x=-5+-2sqrt(5)$

$color(blue)(x~~ -9.472" and "-0.528" to 3 decimal places")$

How do you solve using the completing the square method 0= x^2 + 10x + 50= x^2 + 10x + 5?