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: Understand the given polynomial:
We are given the polynomial: \[ f(t) = t^2 + 4\sqrt{3}t - 15 \] We need to find the zeroes of this polynomial and verify the relationship between the zeroes and the coefficients.

Step 2: Use the quadratic formula to find the zeroes:
The general form of a quadratic equation is: \[ at^2 + bt + c = 0 \] For the given polynomial $f(t) = t^2 + 4\sqrt{3}t - 15$, we have: - $a = 1$ - $b = 4\sqrt{3}$ - $c = -15$ The quadratic formula to find the roots (zeroes) of the equation $at^2 + bt + c = 0$ is: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the values of $a$, $b$, and $c$ into the formula: \[ t = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(-15)}}{2(1)} \] Simplify the terms: \[ t = \frac{-4\sqrt{3} \pm \sqrt{48 + 60}}{2} \] \[ t = \frac{-4\sqrt{3} \pm \sqrt{108}}{2} \] \[ t = \frac{-4\sqrt{3} \pm 2\sqrt{27}}{2} \] \[ t = \frac{-4\sqrt{3} \pm 6\sqrt{3}}{2} \] Now, simplify further: \[ t = \frac{-4\sqrt{3} + 6\sqrt{3}}{2} \quad \text{or} \quad t = \frac{-4\sqrt{3} - 6\sqrt{3}}{2} \] \[ t = \frac{2\sqrt{3}}{2} = \sqrt{3} \quad \text{or} \quad t = \frac{-10\sqrt{3}}{2} = -5\sqrt{3} \] So, the zeroes of the polynomial are: \[ t_1 = \sqrt{3} \quad \text{and} \quad t_2 = -5\sqrt{3} \]

Step 3: Verify the relationship between the zeroes and the coefficients:
From Vieta's formulas, for a quadratic equation $at^2 + bt + c = 0$, the sum and product of the roots are related to the coefficients as follows: - The sum of the zeroes $t_1 + t_2 = -\frac{b}{a}$ - The product of the zeroes $t_1 \times t_2 = \frac{c}{a}$ For the given polynomial $f(t) = t^2 + 4\sqrt{3}t - 15$: - The sum of the zeroes is: \[ t_1 + t_2 = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} \] - According to Vieta’s formula, the sum of the zeroes should be: \[ -\frac{b}{a} = -\frac{4\sqrt{3}}{1} = -4\sqrt{3} \] Thus, the sum of the zeroes is correct. - The product of the zeroes is: \[ t_1 \times t_2 = \sqrt{3} \times (-5\sqrt{3}) = -5 \times 3 = -15 \] - According to Vieta’s formula, the product of the zeroes should be: \[ \frac{c}{a} = \frac{-15}{1} = -15 \] Thus, the product of the zeroes is also correct. Therefore, the relationship between the zeroes and the coefficients is verified.

Conclusion:
The zeroes of the polynomial $f(t) = t^2 + 4\sqrt{3}t - 15$ are $t_1 = \sqrt{3}$ and $t_2 = -5\sqrt{3}$. The sum and product of the zeroes are consistent with the coefficients of the polynomial.