How do you complete the square to solve #2x² + 3x -5=0#?

2 Answers
Jul 2, 2015

Answer:

#x = 1# or #x = -5/2#

Explanation:

You follow a series of steps to complete the square.

Step 1. Write your equation in standard form.

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

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

Add #5# to each side of the equation.

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

#2x^2+3x = 5#

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

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

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

#(3/2)^2/4 = (9/4)/4 = 9/16#

Step 5. Add the result to each side.

#x^2 +3/2x + 9/16 =5/2 + 9/16#

#x^2 +3/2x + 9/16= 40/16 + 9/16#

#x^2 +3/2x + 9/16 =49/16#

Step 6. Take the square root of each side.

#x+3/4 = ±7/4 #

#x_1 + 3/4 = +7/4#

#x_1 = 7/4 – 3/4 = (7-3)/4 = 4/4 = 1#

#x_1 = 1#

#x_2 + 3/4 = - 7/4#

#x_2 = -7/4 – 3/4 = (-7-3)/4 = (-10)/4 = -5/2#

So #x = 1# or #x = -5/2#

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

(a) #x = 1#

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

(b) #x = -5/2#

#2x^2 + 3x -5 = 2(-5/2)^2 + 3(-5/2) -5 = 2(25/4) – 15/2 -5 = 25/2 -15/2-10/2 = (25-15-10)/2 = 0#

Jul 2, 2015

Answer:

Solve: #y = 2x^2 + 3x - 5 = 0#

Explanation:

To solve this type of quadratic equations, use the Shortcut. It can save you a lot of work and time.
(a + b + c = 0) -> 2 real roots: 1 and #(c/a = - 5/2)#