How do you solve #3x^2 + 17x -6 = 0# by completing the square?

Question

Isaiah Anthony · Accepted Answer

#color(blue)(x = 1/3, -6#

#3x2 + 17x - 6 # 0#

Cancel3(x^2 + (17/3)x -2) to get #cancel3

Using #x^2 + color(purple)((2 * (17/6) x) ) = 2#, the middle term #2xy# is created.

To create a perfect square, add #color(green)( (17/6)^2#) to both sides of the L H S.

#x^2 + (2 * (17/6)x + (17/6)^2 = 2 + (17/6)^2 = 361/36#

(19/6)^2 = #(x+(17/6))^2#

19/6# #x + 17/6 = color(red)( +-)

19/6 - 17/6# #x = color(red)( +-)

Blue #color (x = 1/3, -6#)

Ethan Adkins · Answer

undefined To solve the quadratic equation 3x^2 + 17x - 6 = 0 by completing the square, follow these steps:

1. Move the constant term to the other side:
   3x^2 + 17x = 6

2. Divide the coefficient of the x^2 term (3) from both sides to make the coefficient of x^2 equal to 1:
   x^2 + (17/3)x = 2

3. Take half of the coefficient of x, square it, and add it to both sides of the equation:
   x^2 + (17/3)x + (17/6)^2 = 2 + (17/6)^2

4. Rewrite the left side as a squared binomial:
   (x + 17/6)^2 = 2 + 289/36

5. Simplify the right side:
   (x + 17/6)^2 = 72/36 + 289/36
   (x + 17/6)^2 = 361/36

6. Take the square root of both sides:
   x + 17/6 = ±√(361/36)

7. Solve for x:
   x = -17/6 ± √(361/36)
   x = -17/6 ± 19/6

8. Simplify:
   x = (-17 ± 19)/6

9. Find the two solutions:
   x₁ = (-17 + 19)/6 = 2/6 = 1/3
   x₂ = (-17 - 19)/6 = -36/6 = -6

So, the solutions to the equation 3x^2 + 17x - 6 = 0 by completing the square are x = 1/3 and x = -6.