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

2 Answers
Aug 3, 2016

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

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

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