y^2+18y+32=0 how do u solve the equation?

Question

how do you solve the equation

Answer 1

Hi Michelle -

Since this equation is of the form: ay^2 + by+ c = 0, it is a quadratic equation, and we can use the quadratic formula to solve it:

y = (-b +- (b^2 - 4ac)^(1/2))/2a  (yes, I know it looks weird, but only because of how you have to type it here).  Note that a value^(1/2) or a value to the 1/2 power is the same thing as the square trot of that value.

So we have: 

y^2 + 18y + 32 = 0, and comparing to ay^2 + by+ c = 0, we have in our case:

a = 1, b = 18, and c = 32

Plugging in these values to the above quadratic formula, we get:

y = (-18 +- (18^2 -4•1•32)^(1/2))/2•1

Simplifying, we get:

y = (-18 +- (324 - 128)^(1/2))/2

Simplifying some more, we get:

y = (-18 +- (196^(1/2)))/2

Then:

y = (-18 +- 14)/2  (since the square root of 196 is 14)

Then, we have to solutions for y:

y = (-18 + 14)/2 and y = (-18 - 14)/2

Or:

y = -4/2 and y = -32/2

Finally:

y = -2 and y = -16

Hope this helps!

Mike T.

 

Answer 2

y² + 18y + 32 = 0

 

Solve by Completing the Square

 

y^{2 + 18y + 32 -32 = -32}

 

y²  + 18y = -32

 

Middle term is 18 so; 18/2 = 9 finds the end term to complete the square which is 9²  or 81.

 

y² + 18y + 81 = 81 - 32

 

(y + 9)² = 49

 

Taking the square root of both sides:

 

y + 9 = ±7

 

This give two solutions; y = -2, -16

Answer 3

Hi Michelle . first you factor this eqn , if you look at the eqn , you see that 32 has factors 2 and 16 , so the 

solved eqn is ( y + 2) (y+16 ) =0, implying y=-2 or y=-16