How do you solve #y^2 + 10y + 35 = 0# by completing the square?

2 Answers
Aug 3, 2016

Answer:

#y=-5+sqrt(-10)#
#y=-5-sqrt(-10)#

Explanation:

#y^2+10y+35=0#
or
#y^2+10y+25+10=0#
or
#y^2+2(y)5+5^2+10=0#
or
#(y+5)^2=-10#
or
#y+5=+-sqrt(-10)#
or
#y=-5+-sqrt(-10)#
or
#y=-5+sqrt(-10)#========Ans #1#
or
#y=-5-sqrt(-10)#========Ans #2#

Aug 3, 2016

Answer:

#y=-5+-(sqrt(-10)) i#

Explanation:

This is a quadratic in #y# instead of in #x#. This means that the plot will be rotated #90^0# and be of shape #sub# or #sup#.

Set as #" "x=y^2+10y+35#

#color(blue)("Step 1")#

Write as:#" "x=(y^2+10y)+35+k#

The #k# will be needed to correct an error that this approach introduces.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Step 2")#

Take the index (power) outside the brackets
#x=(y+10y)^2+35+k#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Discard the #y# from #10y#
#x=(y+10)^2+35+k#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#
Halve the 10
#x=(y+5)^2+35+k#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")#
Now we need to correct for the error. The 5 in #(y+5)^2# produces #5^2# which is a value that is not in the original equation

So #5^2+k=0 => k=-25# giving:

#color(brown)(x=(y+5)^2+35+k)color(blue)(" "->" "x=(y+5)^2+35-25#

#color(green)(x=(y+5)^2+10) " "larr" Completed the square"#

This is also known as the Vertex Form Equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving for "(y+5)^2+10=0)#

#(y+5)^2=-10#

Square root both sides

#y+5=+-sqrt(-10) #

As soon as you see a root of a negative it means Complex Numbers are involved. In other words; the plot does not cross the y-axis in this case. If the graph was of general shape #uu" or "nn# it would not cross the x-axis. This graph is of shape #sub#.

#y=-5+-(sqrt(-10)) i#

Tony B