How do you solve using the completing the square method #y^2+4y-45=0#?

Question

Kennedy Adams · Accepted Answer

#y=-9 or y=5#

To complete the square we use #1/2# of the #y# coefficient as follows:

# y^2 + 4y -45 = 0 # # :. (y^2 + 1/2(4))^2 - (1/2(4))^2 - 45 = 0# # :. (y^2 + 2)^2 - (2)^2 - 45 = 0# # :. (y^2 + 2)^2 - 4 - 45 = 0# # :. (y^2 + 2)^2 - 49 = 0# # :. (y^2 + 2)^2 = 49 #

We can now take the square root remembering that when we do so we get two solutions as #a^2 = b <=> a = +- sqrtb #

# :. y^2 + 2 = +-7 # # :. y^2 = -2+-7 #

obtaining the two answers:

#y=-9 or y=5#

Savannah Anderson · Answer

undefined To solve the quadratic equation y^2 + 4y - 45 = 0 using the completing the square method, follow these steps:

1. Move the constant term to the other side of the equation:
   y^2 + 4y = 45

2. Take half of the coefficient of y (which is 4) and square it:
   (4/2)^2 = 4

3. Add the square of half the coefficient of y to both sides of the equation:
   y^2 + 4y + 4 = 45 + 4
   y^2 + 4y + 4 = 49

4. Rewrite the left side as a perfect square:
   (y + 2)^2 = 49

5. Take the square root of both sides:
   y + 2 = ±√49
   y + 2 = ±7

6. Solve for y:
   y = -2 ± 7
   y = -2 + 7 or y = -2 - 7
   y = 5 or y = -9

So, the solutions to the equation y^2 + 4y - 45 = 0 are y = 5 and y = -9.