Question

Find the number that must divide each term in the equation so that the equation can be solved by the method of completing the square: 6x^2-9x=19

a. 6
b. -9
c. x
d. 19

Answer 1

Thanks @wolf1728,
I think the answer is 6.

Answer 2

Is that right? I think it's 6 but I'm not sure!

Answer 3

6x^2-9x=19
To complete the square:
Keep all terms to the left and
move the "non X" term to the right:
6x^2 -9x = 19

Divide the equation by the coefficient of X² which in this case is 6
x² -1.5x = 3.16666666

Now for "completing the square" we:
take the coefficient of X which is -1.5
divide that by 2
-.75
square that number
.5625
then add it to both sides of the equation
x² -1.5x +.5625= 3.16666666 +.5625
Take the square root of BOTH sides of the equation
x-.75 = 1.931105036
x = 2.681105036

x-.75 = -1.931105036
x = -1.181105036

Well I completed the square but I don't know which of the options to choose.
x +.75= 1.931105036

Answer 4

Thank you so much!!!!!!

Answer 5

Sheesh it's just that simple? Divide by 6?
Well, glad to help out.
:-)

Answer 6

To complete the square
\[ (x+p)^2 = x^2 + 2px + p^2\]

You have
\[6x^2 - 9x =19\]


****You have to get rid of the coefficient in front of x to fit the form, so divide by 6****

and add p^2 to each side
\[x^2 + \left( -\frac{ 9}{6}x \right) + (p^2) =-\frac{19}{6} +( p^2 )\]

Solve for p

\[ 2p = -\frac{9}{6} \]
\[ p = -\frac{9}{12} = -\frac{3}{4} \]
\[ p^2 = \frac{9}{16} \]
\[(x-\frac{3}{4})^2 = x^2 -\frac{9}{12}x + \left( \frac{9}{16} \right) = \frac{19}{6} + \left( \frac{9}{16} \right) \]

\[(x-\frac{3}{4})^2 = \frac{179}{48} \]

\[x = \frac{3}{4} ± \sqrt{ \frac {179}{48}}\]