MHB Find zeros of polynomial and factor it out, find the reals and complex numbers

[datafiend]

Hi all,
$$f(x) = 3x^2+2x+10$$

I recognized that this a quadratic and used the quadratic formula. I came up with $$-1/3+-\sqrt{29}/3$$.

But the answer has a $$i$$ for imaginary. When I was under the \sqrt{116}, I broke that down, but didn't realize there would be an $$i$$

Can someone explain that one to me?

Thanks

 

[evinda]

Hi all,
$$f(x) = 3x^2+2x+10$$

I recognized that this a quadratic and used the quadratic formula. I came up with $$-1/3+-\sqrt{29}/3$$.

But the answer has a $$i$$ for imaginary. When I was under the \sqrt{116}, I broke that down, but didn't realize there would be an $$i$$

Can someone explain that one to me?

Thanks

Hi!

$$\Delta=b^2-4ac=2^2-4 \cdot 3 \cdot 10=4-120=-116=116i^2$$

$$x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-2 \pm \sqrt{116i^2}}{6}=\frac{-2 \pm 2 i \sqrt{29}}{6}=\frac{-1 \pm i \sqrt{29}}{3}$$

So, $x_1=\frac{-1+ i \sqrt{29}}{3}$ and $x_2=\frac{-1- i \sqrt{29}}{3}$.

 

[mathbalarka]

$i$ is just a name for $\sqrt{-1}$, the imaginary unit.

The quadratic is $3x^2 + 2x + 10 = 0$. Multiply both sides by $4\cdot 3 = 12$ to get

$$4 \cdot 3^2 x^2 + 4 \cdot 3 \cdot 2x + 120 = (2 \cdot 3 \cdot x)^2 + 2 \cdot (2 \cdot 3) \cdot (2) x + (2)^2 + \left ( -4 + 120 \right)$$

By completing the square, one gets

$$(2 \cdot 3 x + 2)^2 + 116 = 0$$

And solving for $x$ results

$$x = \frac{-2 \pm \color{red}{\sqrt{-116}}}{6}$$

But we know that $\sqrt{ab} = \sqrt{a}\sqrt{b}$, thus $\sqrt{-116} = \sqrt{-1}\cdot \sqrt{166} = \sqrt{-1} \cdot 2 \cdot \sqrt{29} = \boxed{i2\sqrt{29}}$ which is the desired numerator.

 

[MarkFL]

If we complete the square to write the function in vertex form as follows, we find:

$$f(x) = 3x^2+2x+10=3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+10-3\cdot\frac{1}{9}=3\left(x-\left(-\frac{1}{3}\right)\right)^2+\frac{29}{3}$$

We can then see that:

$$f_{\min}=f\left(-\frac{1}{3}\right)=\frac{29}{3}>0$$

So, we see that for any real value of $x$, the given function is greater than zero, and thus has no real roots. Thus, we should expect the roots to be complex.