Maths: Quadratics Help?

Question

I need help to solve this quadratic:

3x^2 + 5x - 3 = 0


I know that you have to used the compleating the square method and I also know the answer:

x= -5 plus minus the square root of 61 over 6

How would I solve it?

Answer 1

The answer you have given suggests that the quadratic formula should be used.

x = (-b +/- sqrt(b^2 - 4ac))/2a

Where your quadratic is in the form
ax^2 + bx + c = 0

So for you
a = 3, b=5 c = -3.

Plug these numbers into the formula and you should be able to work them through to your answer.
Good Luck :)

Answer 2

Quadratic Formula

x = - b ± √b² - 4ac / 2a

3x² + 5x - 3 = 0

Let

a = 3

b = 5

c = - 3

- - - - - - - -

x = - 5 ± √(5)² - 4(3)(- 3) / 2(3)

x = - 5 ± √25 - (- 36) / 6

x = - 5 ± √25 + 36 / 6

x = - 5 ± √61 / 6

x = - 5 ± 7.810249676 / 6

- - - - - - - -

Solving for +

x = - 5 + 7.810249676 / 6

x = 2.810249676 / 6

x = 0.468374946

- - - - - -

Solving for -

x = - 5 - 7.810249676 / 6

x = - 12.81024968 / 6

x = - 2.135041613

- - - - - - -s-

Answer 3

Using your method of completing the square gives:-
3(x² + (5/3).x - 1) = 0
[ x² + (5/3).x + 25/36 ] - 25/36 - 1 = 0
[x + (5/6)]² = 61/36
x + 5/6 = ± (1/6).√61
x = (- 5/6) ± (1/6).√61
x = (-1/6).[ 5 ±√61]

Answer 4

use the quadratic formula:
x=[-b+/-√(b^2-4ac)]/2a
for ax^2+bx+c=0
x= [-5+/-√(25+36)]/6
or
x^2+(5/3)x-1=0
x^2+(5/3)x+(5/6)^2-(5/6)^2-1=0
(x+ 5/6)^2=61/36
x+ 5/6 =+/- √(61/36)
x= -5/6 +/- (√61)/6