How do you solve # 2x^2-3x-14=0# by completing the square?

Question

Hazel Anglin · Accepted Answer

#x=7/2" "# or #" "x=-2#

The squares identity difference can be expressed as follows:

#A^2-B^2 = (A-B)(A+B)#

We will use this with #A=(4x-3)# and #B=11# later.

#color(white)()# Given:

#2x^2-3x-14 = 0#

To avoid fractions as much as possible, let us premultiply this quadratic by #8 = 2*2^2#. One factor of #2# makes the leading term a perfect square, then the other #2^2# deals with the #2# denominator in "#b/(2a)#" which would otherwise cause us to have to work with #1/2#'s.

So:

#0 = 8(2x^2-3x-14)#

#color(white)(0) = 16x^2-24x-112#

#color(white)(0) = (4x)^2-2(4x)(3)+9-121#

#color(white)(0) = (4x-3)^2-11^2#

#color(white)(0) = ((4x-3)-11)((4x-3)+11)#

#color(white)(0) = (4x-14)(4x+8)#

#color(white)(0) = (2(2x-7))(4(x+2))#

#color(white)(0) = 8(2x-7)(x+2)#

Hence:

#x=7/2" "# or #" "x=-2#

Genesis Allen · Answer

undefined To solve the quadratic equation $ 2x^2 - 3x - 14 = 0 $ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $ 2x^2 - 3x = 14 $

2. Divide all terms by the coefficient of $ x^2 $ (in this case, 2):
   $ x^2 - \frac{3}{2}x = 7 $

3. Take half of the coefficient of $ x $ (in this case, $ -\frac{3}{2} $) and square it:
   $ \left(-\frac{3}{4}\right)^2 = \frac{9}{16} $

4. Add and subtract the result from step 3 to both sides of the equation:
   $ x^2 - \frac{3}{2}x + \frac{9}{16} = 7 + \frac{9}{16} $

5. Rewrite the left side as a perfect square:
   $ \left(x - \frac{3}{4}\right)^2 = \frac{112}{16} + \frac{9}{16} $

6. Simplify the right side:
   $ \left(x - \frac{3}{4}\right)^2 = \frac{121}{16} $

7. Take the square root of both sides:
   $ x - \frac{3}{4} = \pm \frac{11}{4} $

8. Solve for $ x $:
   $ x = \frac{3}{4} \pm \frac{11}{4} $

9. Simplify:
   $ x = \frac{14}{4} $ or $ x = -\frac{8}{4} $

10. Further simplify:
    $ x = \frac{7}{2} $ or $ x = -2 $

Therefore, the solutions to the equation $ 2x^2 - 3x - 14 = 0 $ are $ x = \frac{7}{2} $ and $ x = -2 $.