Question

a circle has the equation below. *
what is the radius of this circle?

$$x^2 + y^2 - x + 10y + 13 = 0$$

your answer

how many and which types of roots does the function $f(x)$ have? *
refer to the image below:

$$f(x) = 18x^4 + 73x^2 - 36$$

• 4 real roots
• 1 real root and 3 imaginary roots
• 3 real roots and 1 imaginary root
• 2 real roots and 2 imaginary roots

Explanation:

First Sub - Question (Radius of the Circle)

Step1: Rewrite the circle equation

The general form of a circle's equation is $x^{2}+y^{2}+Dx + Ey+F = 0$, and we can convert it to the standard form $(x - h)^{2}+(y - k)^{2}=r^{2}$ by completing the square.
For the given equation $x^{2}+y^{2}-x + 10y+13 = 0$, we group the $x$ - terms and $y$ - terms:
$(x^{2}-x)+(y^{2}+10y)=- 13$

Step2: Complete the square for $x$ - terms

For the $x$ - terms $x^{2}-x$, we use the formula $a^{2}-2ab + b^{2}=(a - b)^{2}$. Here, $a = x$ and $2ab=x$, so $b=\frac{1}{2}$. Then $x^{2}-x=(x-\frac{1}{2})^{2}-\frac{1}{4}$

Step3: Complete the square for $y$ - terms

For the $y$ - terms $y^{2}+10y$, we use the formula $a^{2}+2ab + b^{2}=(a + b)^{2}$. Here, $a = y$ and $2ab = 10y$, so $b = 5$. Then $y^{2}+10y=(y + 5)^{2}-25$

Step4: Substitute back and simplify

Substitute the completed - square forms back into the equation:
$(x-\frac{1}{2})^{2}-\frac{1}{4}+(y + 5)^{2}-25=-13$
$(x-\frac{1}{2})^{2}+(y + 5)^{2}=-13+\frac{1}{4}+25$
$(x-\frac{1}{2})^{2}+(y + 5)^{2}=\frac{- 52 + 1+100}{4}$
$(x-\frac{1}{2})^{2}+(y + 5)^{2}=\frac{49}{4}$
The standard form of a circle is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $r$ is the radius. So, $r^{2}=\frac{49}{4}$, and $r=\frac{7}{2}$ (since radius is positive)

Second Sub - Question (Roots of the Function)

Let $t=x^{2}$, then the function $f(x)=18x^{4}+73x^{2}-36$ can be rewritten as a quadratic function: $g(t)=18t^{2}+73t - 36$, where $t\geq0$ (because $t = x^{2}$)

Step1: Solve the quadratic equation for $t$

We use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$. Here, $a = 18$, $b = 73$, $c=-36$
First, calculate the discriminant $\Delta=b^{2}-4ac=(73)^{2}-4\times18\times(-36)=5329 + 2592=7921$
Then $t=\frac{-73\pm\sqrt{7921}}{2\times18}=\frac{-73\pm89}{36}$
We get two solutions for $t$:
$t_{1}=\frac{-73 + 89}{36}=\frac{16}{36}=\frac{4}{9}$ and $t_{2}=\frac{-73 - 89}{36}=\frac{-162}{36}=-\frac{9}{2}$

Step2: Analyze the solutions for $x$

Since $t=x^{2}\geq0$, we discard the solution $t =-\frac{9}{2}$ (because $x^{2}$ cannot be negative).
For $t=\frac{4}{9}$, we have $x^{2}=\frac{4}{9}$, which gives $x=\pm\frac{2}{3}$ (two real roots).
The equation $x^{2}=-\frac{9}{2}$ has solutions $x=\pm\frac{3i}{\sqrt{2}}$ (two imaginary roots). So the function $f(x)$ has 2 real roots and 2 imaginary roots.

Answer:

$\frac{7}{2}$