how do u do completing the sqare?

Question

i want to do this prob.
3x^2 + 5x - 4
how do u do it?

Answer 1

3x^2 + 5x - 4 = 0

add 4 for both sides
3x^2 + 5x = 4

divide both sides by 3
x^2 + 5/3x = 4/3

prepare for completing square
x^2 + 5/3x + ? = 4/3 + ?

use (b/2)^2
x^2 + 5/3x + ( (5/3) / 2 )^2 = 4/3 + ( (5/3) / 2 )^2

multiply them out
x^2 + 5/3x + 25/36 = 4/3 + 25/36

combine like terms
x^2 + 5/3x + 25/36 = 73/36

factor
(x+5/6)^2 = 73/36

take a square root
x + 5/3 = +/- sqr(73)/6

subtrac 5/3
x = -5/3 +/- sqr(73)/6

hope this helps.

Answer 2

Bring your -4, over to the other side of the equation

Divide everything in your equation by 3.

x^2 + 5/3 x = 4/3

take the 5/3 term, half it, square it & add it (to both sides of the equation!!)

x^2 + 5/3 x + 25/9 = 4/3 + 25/9

[ x + 5/6 ]^2 = 4/3 + 25/9

From here, simply add the fractions on the right.

Take the square root of both sides.

& then bring the 5/6 fraction over to the other side of the equation and you have your answer.

Hope that helps!! :)

Don't forget that your square root term can be negative or positive. Include both answers.

Answer 3

= 3 [ x² + (5/3) x - 4/3 ]

= 3[ (x² + 5/3 x + 25 / 36) - 25/36 - 4/3]

= 3[( x + 5/6)² - 25/36 - 48/36]

= 3[ (x + 5/6)² - 73/36]

Square has been completed as requested.
There is no request to solve an equation so no equation was solved!

Answer 4

first make the coefficient of x^2 unity by dividing by 3

then take half the coefficient of x with the sign that is half of (5/3)
that is (5/6)

square (5/6) and add it to both sides and the express the L.H.S as a square

then take the square root

try the following page

Answer 5

3x ^2 + 5x - 4

= 3 ( x^2 + 5/3 x - 4/3)

= 3 ( x^2 + 2 ( 5/6)x + (5/6)^2 - (5/6)^2 - 4/3)

= 3 ( ( x + 5/6) ^2 - 73/36)

= 3 (( x + 5/6) ^2 - (sq rt 73)/6)

= 3 ( x + 5/6 + (sq. rt 73) /6) ( x + 5/6 - (sq. rt. 73)/6)