How do I complete the square for 4x^2+ 8x-5= 0?

Question

asked Aug 10, 2022 126k views

2 Answers

Cathyann · Answer 1 · 2022-08-11T21:36:02+0000

Answer:

x =
(1)/(2) , -
(5)/(2)

Explanation:

This is not so fun!

$4x^(2) + 8x - 5 = 0\\$

Step 1: Move the 5 to the right side

4x^(2) +8x = 5

Step 2: Divide thru by 4

x^(2) + 2x = (5)/(4)

Step 3: Divide 2 (from 2x) by 2 and square the result and add that to both sides of the equation.

2/2 = 1 1 squared = 1

x^(2) + 2x + 1 = (5)/(4) +1 = (9)/(4)

Step 4: Factor the left side

(x + 1)^(2) = (9)/(4)

Step 5: Take the square root of both sides

x + 1 = ±
$\sqrt{(9)/(4) }$ = ±
(3)/(2)

Step 6: Move the 1 to the right side and add the results

x = -1 ±
(3)/(2) = -1 +
(3)/(2) or -1 -
(3)/(2)

=
(1)/(2) or -
(5)/(2)

And bingo, you have the solutions by completing the square. Lot of fun, huh.