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

[datafiend]

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

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

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

Can someone explain that one to me?

Thanks

 

[evinda]

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

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

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

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:

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

We can then see that:

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

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.

 

[CellCoree]

for sharing your findings with us. It seems like you have correctly found the zeros of the polynomial using the quadratic formula. The presence of the imaginary number, i, in the solution is due to the fact that the discriminant (the part under the square root sign) is negative. This indicates that the solutions are complex numbers.

In order to find the real and complex numbers, we can rewrite the polynomial as a product of two factors: f(x) = (3x+5)(x+2). This tells us that the zeros of the polynomial are x = -5/3 and x = -2. These are the real numbers. However, when we plug these values back into the original polynomial, we get a complex result. This is because the complex numbers are needed to complete the square and factor out the polynomial.

In summary, the complex numbers in the solution indicate that the polynomial has complex roots, but we can still find the real numbers by factoring it out. I hope this explanation helps. Keep up the good work!