How do you solve #2y ^ { 2} - 5x - 12= 0#?

1 Answer
Dec 5, 2016

Answer:

#y=+-sqrt(5/2x+6)#

Explanation:

Taking it 1 step at a time so that you can see better what is going on

Using first principles (short cut method taken from this approach)

#color(purple)("Step 1")#

Add #color(red)(5x)# to both sides

#color(blue)(2y^2-5x-12=0" "->" "2y^2color(red)(+5x)-5x-12=0color(red)(+5x))#

#" "2y^2" "+0" "-12=5x#

#" "2y^2-12=5x#

...................................................................................................

#color(purple)("Step 2")#

Add #color(red)(12)# to both sides

#color(blue)(2y^2-12=5x" "->" "2y^2color(red)(+12)-12=5xcolor(red)(+12) )#

#" "2y^2" "+0" "=5x+12#

#" "2y^2=5x+12#

...................................................................................................

#color(purple)("Step 3")#

Divide both sides by 2 (same as #color(red)(xx1/2)#)

#color(blue)(2y^2=5x+12" "->" "2/(color(red)(2))y^2=5/(color(red)(2))x+12/(color(red)(2)) #

But #2/2=1# giving:#" "y^2=5/2x+6#

...................................................................................................

#color(purple)("Step 4")#

Square root both sides:#" "sqrt(y^2)=sqrt(5/2x+6)#

#" "y=+-sqrt(5/2x+6)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(purple)(ul("Foot note"))#

Using an example #(-2)xx(-2)=+4#
#" "(+2)xx(+2) = +4#

As both #(-2)xx(-2)" and "(+2)xx(+2)# both equal 4 then

#sqrt(4) = "plus or minus "2 -> +-2#

Which is why we end up with #y=+-sqrt(5/2x+6)#