Help please!!!?

Question

How do I solve by completing the square?

6x^2 -7 = -11x

What do I need to do?

Answer 1

First get everything onto one side: 6x^2+11x-7=0
Next factor out the 6: 6(x^2+11/6x-7/6)=0
so x^2+11/6x-7/6=0
now add half of 11/6 to x and square:
(x+11/12)^2--but the constant term would be 121/144 but needs to be -7/6=-168/144
Therefore, subtract 289/144 to get
(x+11/12)^2-289/144=0
next add 289/144 to both sides to get:
(x+11/12)^2=289/144
now take the square root of both sides:
x+11/12=+-17/12
so x=-11/12+-17/12

Answer 2

Add 11x to both sides

6x² + 11x - 7 = 0

Completing the square:

Divide both sides by 6 to make the coefficient of x² be 1.
x² + (11/6)x - (7/6) = 0

Half the coefficient of x is 11/12

(x + 11/12)² = x² + (11/6)x +(11/12)²

So the first two terms are what we want, we just need to deal with the last term.

(x + 11/12)² - (11/12)² is x² + (11/6)x

so (x + 11/12)² - (11/12)² - (7/6) = x² + (11/6)x - (7/6)

Since x² + (11/6)x - (7/6) = 0
then (x + 11/12)² - (11/12)² - (7/6) = 0

(x + 11/12)² - (289/144) = 0
(x + 11/12)² = (289/144)

Square root both sides:
x + 11/12 = ± √(289/144)
x + 11/12 = ± (17/12)
x = -11/12 ± (17/12)
x = (-11 ± 17)/12
x = -7/3 or x = 1/2


We can check using the quadratic formula:

(Given:
ax² + bx +c = 0
then x = - b ± √(b² - 4ac)/2a)

x = (-11 ±√((11)² - 4(6)(-7)))/2(6)
x = (-11 ±√((11)² - 4(6)(-7)))/2(6)
x = (-11 ±√(289))/12
x = (-11 ±17)/12
x = -7/3 or x = 1/2

Answer 3

6x^2 + 11x = 7
x^2 + 11/6x = 7/6

You complete the square by taking the B term, dividing it in half, and squaring it.

11/6 / 2 = 11/12
(11/12)^2 = 121/144

So the equation must look like

x^2 + (11/6) + 121/144 = 7/6 + 121/144
(x + 11/12)^2 = 289/144
(x + 11/12) = +/- 17/12
x = 6/12 or x = -28/12
x = 1/2 or x = -2-1/3

Answer 4

6x^2 -7 + 21/4 = -23/4

6x^2 = 1

6x=1

x = 1/6