How do you complete the square to solve #0=5x^2 + 2x - 3#?

Question

2592 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-03T14:45:40

#x = 3/5# or #x = -1#

Step 1. Write your equation in standard form.

#5x^2 + 2x -3 = 0#

Step 2. Move the constant to the right hand side of the equation.

Add #3# to each side .

#5x^2+2x -3 +3 = 0+3#

#5x^2+2x = 3#

Step 3. Divide both sides of the equation by the coefficient of #x^2#.

Divide both sides by 5.

#x^2 +2/5x =3/5#

Step 4. Square the coefficient of x and divide by 4.

#(2/5)^2/4 = (4/25)/4 = 1/25#

Step 5. Add the result to each side.

#x^2 +2/5x + 1/25 =3/5 + 1/25#

#x^2 +2/5x + 1/25= 15/25 + 1/25#

#x^2 +2/5x + 1/25 =16/25#

Step 6. Take the square root of each side.

#x+1/5 = ±4/5 #

Case 1

#x_1 + 1/5 = +4/5#

#x_1 = 4/5-1/5 = (4-1)/5#

#x_1 = 3/5#

Case 2

#x_2 + 1/5 = -4/5#

#x_2 = -4/5-1/5 = (-4-1)/5 = (-5)/5#

#x_2 = -1#

So #x = 3/5# or #x = -1#

Check: Substitute the values of #x# back into the quadratic.

(a) #x = 3/5#

#5x^2 + 2x -3 = 5(3/5)^2 + 2(3/5) -3 = 5(9/25) + 6/5 -3 = 9/5 +6/5 -15/5 = (9+6-15)/5 = 0#.

(b) #x = -1#

#5x^2 + 2x -3 = 5(-1)^2 + 2(-1) -3 = 5(1) – 2 -3 = 5-2-3 = 0#