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

Answer 1

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

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.

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

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Answer 2

First we need to remember the formula:

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.)

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Answer 3

To solve the quadratic equation -3n^2 + 5n - 2 = 0 using the quadratic formula, which is (n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = -3), (b = 5), and (c = -2):

  1. Substitute the values of (a), (b), and (c) into the quadratic formula.
  2. Calculate the discriminant, which is (b^2 - 4ac).
  3. If the discriminant is positive, you will have two real solutions. If it's zero, there will be one real solution (a repeated root). If it's negative, there will be no real solutions (two complex solutions).
  4. Plug the values of (a), (b), and the discriminant into the quadratic formula and solve for (n).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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