Question

Find the zeroes of the polynomial $f(t) = t^2 + 4\sqrt{3}t - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Answer 1

Step 1: Use the quadratic formula The polynomial is:
$f(t) = t^2 + 4\sqrt{3}t - 15.$
Using the quadratic formula:
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$where $a = 1$, $b = 4\sqrt{3}$, and $c = -15$. Substitute:
$t = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(-15)}}{2(1)} = \frac{-4\sqrt{3} \pm \sqrt{48 + 60}}{2}.$
Simplify:
$t = \frac{-4\sqrt{3} \pm \sqrt{108}}{2} = \frac{-4\sqrt{3} \pm 6\sqrt{3}}{2}.$
Separate the roots:
$t_1 = \frac{-4\sqrt{3} + 6\sqrt{3}}{2} = \sqrt{3}, \quad t_2 = \frac{-4\sqrt{3} - 6\sqrt{3}}{2} = -5\sqrt{3}.$
Step 2: Verify the relationships
Sum of zeroes: $t_1 + t_2 = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} = -\frac{b}{a}$.
Product of zeroes: $t_1 \cdot t_2 = (\sqrt{3})(-5\sqrt{3}) = -15 = \frac{c}{a}$.
Correct Answer: Zeroes are $\sqrt{3}$ and $-5\sqrt{3}$, and the relationships are verified.

Answer 2

Step 1: Use the quadratic formula The polynomial is:
$f(t) = t^2 + 4\sqrt{3}t - 15.$
Using the quadratic formula:
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$where $a = 1$, $b = 4\sqrt{3}$, and $c = -15$. Substitute:
$t = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(-15)}}{2(1)} = \frac{-4\sqrt{3} \pm \sqrt{48 + 60}}{2}.$
Simplify:
$t = \frac{-4\sqrt{3} \pm \sqrt{108}}{2} = \frac{-4\sqrt{3} \pm 6\sqrt{3}}{2}.$
Separate the roots:
$t_1 = \frac{-4\sqrt{3} + 6\sqrt{3}}{2} = \sqrt{3}, \quad t_2 = \frac{-4\sqrt{3} - 6\sqrt{3}}{2} = -5\sqrt{3}.$
Step 2: Verify the relationships
Sum of zeroes: $t_1 + t_2 = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} = -\frac{b}{a}$.
Product of zeroes: $t_1 \cdot t_2 = (\sqrt{3})(-5\sqrt{3}) = -15 = \frac{c}{a}$.
Correct Answer: Zeroes are $\sqrt{3}$ and $-5\sqrt{3}$, and the relationships are verified.