Square Root Property

[CodySaladbar]

I can't seem to figure out how to solve this equation. Any help please?

3n^2 -n + 5 = 0

The answer should be n = 1/6 (Plus or Minus) √59)/6

I would appreciate it if I can get step by step instructions. Thanks!

 

[MarkFL]

The roots you cite should actually be the complex conjugates \(\displaystyle n=\dfrac{1\pm i\sqrt{59}}{6}\)

We are given:

\(\displaystyle 3n^2-n+5=0\)

I would identify:

\(\displaystyle a=3,\,b=-1\)

1.) Arrange the equation as:

\(\displaystyle 3n^2-n=-5\)

2.) Multiply through by \(\displaystyle 4a=12\)

\(\displaystyle 36n^2-12n=-60\)

3.) Add \(\displaystyle b^2=1\) to both sides:

\(\displaystyle 36n^2-12n+1=-59\)

4.) Write the left side as a square:

\(\displaystyle (6n-1)^2=-59\)

5.) Apply the square root property:

\(\displaystyle 6n-1=\pm\sqrt{-59}=\pm i\sqrt{59}\)

6.) Solve for \(\displaystyle n\):

\(\displaystyle 6n=1\pm i\sqrt{59}\)

\(\displaystyle n=\dfrac{1\pm i\sqrt{59}}{6}\)

 

[JeffM]

I can't seem to figure out how to solve this equation. Any help please?

3n^2 -n + 5 = 0

The answer should be n = 1/6 (Plus or Minus) √59)/6

I would appreciate it if I can get step by step instructions. Thanks!

Mark showed you one way. That way can be generalized into the quadratic formula, which is easier to apply once you have memorized it.

\(\displaystyle ax^2 + bx + c = 0 \implies x = \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}.\) MEMORIZE this quadratic formula.

To apply the quadratic formula to this specific problem

\(\displaystyle Let\ n = x,\ 3 = a,\ b = - 1,\ and\ c = 5 \implies\)

\(\displaystyle n = \dfrac{- (-1) \pm \sqrt{(-1)^2 - (4 * 3 * 5)}}{2 * 3} = \dfrac{1 \pm \sqrt{1 - 60}}{6} = \dfrac{1 \pm \sqrt{-59}}{6} = \dfrac{1 \pm i\sqrt{59}}{6}.\)

Exact same answer as Mark's answer. Easier to do (but it implicitly takes advantage of Mark's logic.)

Now let's check that this answer is correct

\(\displaystyle 3\left(\dfrac{1 + i\sqrt{59}}{6}\right)^2 - \left(\dfrac{1 + i\sqrt{59}}{6}\right) + 5 = \dfrac{3\left\{1^2 + 2(1)\left(i\sqrt{59}\right) + \left(i\sqrt{59}\right)^2\right\}}{6^2} - \dfrac{1 + i\sqrt{59}}{6} + 5 =\)

\(\displaystyle \dfrac{3\left\{1 + \left(2i\sqrt{59}\right) + (-1) * 59\right\}}{36} - \dfrac{1 + i\sqrt{59}}{6} + 5 = \dfrac{1 + \left(2i\sqrt{59}\right) - 59}{12} + \dfrac{-2 - \left(2i\sqrt{59}\right)}{12} + 5 = \dfrac{1 - 59 - 2}{12} + 5 = - 5 + 5 = 0.\)

If one conjugate root is correct the other must be as well. Let's see.

\(\displaystyle 3\left(\dfrac{1 - i\sqrt{59}}{6}\right)^2 - \left(\dfrac{1 - i\sqrt{59}}{6}\right) + 5 = \dfrac{3\left\{1^2 + 2(1)\left(-i\sqrt{59}\right) + \left(-i\sqrt{59}\right)^2\right\}}{6^2} - \dfrac{1 - i\sqrt{59}}{6} + 5 =\)

\(\displaystyle \dfrac{3\left\{1 - \left(2i\sqrt{59}\right) + (-1) * 59\right\}}{36} - \dfrac{1 - i\sqrt{59}}{6} + 5 = \dfrac{1 - \left(2i\sqrt{59}\right) - 59}{12} + \dfrac{-2 + \left(2i\sqrt{59}\right)}{12} + 5 = \dfrac{1 - 59 - 2}{12} + 5 = - 5 + 5 = 0.\)

You can now handle any quadratic under the sun.

 

[MarkFL]

Another method you might not be taught is as follows:

We are given:

\(\displaystyle 3n^2-n+5=0\)

I would identify:

\(\displaystyle a=3,\,b=-1\)

1.) Divide through by \(\displaystyle a=3\)

\(\displaystyle n^2-\dfrac{1}{3}n+\dfrac{5}{3}=0\)

2.) Shift the roots to the right by 1/2 the coefficient of the linear term \(\displaystyle \dfrac{1}{2}\left(-\dfrac{1}{3} \right)=-\dfrac{1}{6}\):

\(\displaystyle \left(n+\dfrac{1}{6} \right)^2-\dfrac{1}{3}\left(n+\dfrac{1}{6} \right)+\dfrac{5}{3}=0\)

3.) Expand, distribute and collect like terms:

\(\displaystyle \left(n^2+\dfrac{1}{3}n+\dfrac{1}{36} \right)-\dfrac{1}{3}\left(n+\frac{1}{6} \right)+\dfrac{5}{3}=0\)

\(\displaystyle n^2+\dfrac{1}{3}n+\dfrac{1}{36}-\dfrac{1}{3}n-\dfrac{1}{18}+\dfrac{5}{3}=0\)

\(\displaystyle n^2=-\dfrac{59}{36}\)

4.) Apply the square root property:

\(\displaystyle n=\pm\sqrt{-\dfrac{59}{36}}=\pm\dfrac{i\sqrt{59}}{6}\)

5.) To get the roots of the original equation we subtract \(\displaystyle -\dfrac{1}{6}\):

\(\displaystyle n=\dfrac{1\pm i\sqrt{59}}{6}\)