How do you solve #-3n^2 + 5n - 2= 0# using the formula?

2 Answers
Jul 1, 2015

Answer:

First we need to remember the formula:

Explanation:

For #ax^2+bx+c=0#, the solutions are given by:

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

In this problem we have #n# instead of #x#, but the formula is the same.

("Minus b, plus or minus the square root of b squared minus 4 a c all over 2a".)

In order to use the formula, we need to identify #a, b, "and " c#

In: #-3n^2 + 5n - 2= 0#, we have:

#color(red) (a=-3)# and #color(blue)(b=5)# and #color(green)(c=-2)#, so we substitute in the formula:

#n = (-color(blue)((5))+-sqrt(color(blue)((5))^2-4color(red)((-3))color(green)((-2))))/(2color(red)((-3))#

You may notice that every number I substituted is in parentheses in this first step. I think that is a good habit to develop for all formulas.

Now we have some simplification/arithmetic to do (I'll remove the colors now)

#n = (-(5)+-sqrt((5)^2-4(-3)(-2)))/(2(-3)#

# = (-5 +- sqrt (25-(-12)(-2)))/(-6)#

# = (-5 +- sqrt (25-(24)))/(-6)#

# = (-5 +- sqrt 1)/(-6)#

# = (-5 +- 1)/(-6)#.

There are two solutions:

One is #(-5+1)/-6# which simplifies to #(-4)/-6 = 2/3#

And the other solution is #(-5-1)/-6# which simplifies to #(-6)/-6 = 1#

note
At the point where we had:
# x = (-5 +- 1)/(-6)#.
We could have made the denominator positive, be doing this:

#(-5 +- 1)/(-6) = ((-1)(-5 +- 1))/6#

Now #-(-5) = 5#, but what about #-(+-1)#?

Remember that writing #+-1# is just a short way of writing the wto numbers #+1# and #-1#, so what we get, in words, is:

The opposite of plus or minus 1, is minus or plus 1. Which is surely the same as plus or minus 1.
So

#(-5 +- 1)/(-6) = ((-1)(-5 +- 1))/6 = (5+-1)/6#.

Finally, notice that when we finish the arithmetic, we get the same answers:

#(5+1)/6 = 6/6=1# and #(5-1)/6 = 4/6 = 2/3#

(We got the same numbers in the opposite order.)

Jul 2, 2015

Answer:

Solve -3x^2 + 5x - 2 = 0

Explanation:

For this type of equation, we don't need a lengthy solving process.
Use the shortcut.
When (a + b + c = 0), one real root is (1) and the other is (c/a = 2/3).

Remind of Shortcut Rule.

  1. When (a + b + c = 0): 2 real roots -> (1) and (c/a)
    Example : #7x^2 - 15x + 8 = 0.#
    (a + b + c = 0)--> 2 real roots: (1) and #(c/a = 8/7)#
  2. When (a - b + c = 0) --> 2 real roots (-1) and (-c/a).
    Example: #19x^2 + 8x - 11 = 0#
    (a - b + c = 0) -> 2 real roots: (-1) and #(-c/a = 11/19)#

The shortcut will save us a lot of work and effort.