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

Step 2: Use the quadratic formula:
The general quadratic equation is \( at^2 + bt + c = 0 \), and the quadratic formula for finding the roots (zeroes) is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the given polynomial \( f(t) = t^2 + 4\sqrt{3}t - 15 \), we have:
- \( a = 1 \) - \( b = 4\sqrt{3} \) - \( c = -15 \)
Substitute these values into the quadratic formula:
\[ t = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(-15)}}{2(1)} \] Simplify the discriminant part:
\[ t = \frac{-4\sqrt{3} \pm \sqrt{16 \times 3 + 60}}{2} \] \[ t = \frac{-4\sqrt{3} \pm \sqrt{48 + 60}}{2} \] \[ t = \frac{-4\sqrt{3} \pm \sqrt{108}}{2} \] \[ t = \frac{-4\sqrt{3} \pm 6\sqrt{3}}{2} \] Now simplify the two cases:
Case 1: \( t = \frac{-4\sqrt{3} + 6\sqrt{3}}{2} \):
\[ t = \frac{2\sqrt{3}}{2} = \sqrt{3} \] Case 2: \( t = \frac{-4\sqrt{3} - 6\sqrt{3}}{2} \):
\[ t = \frac{-10\sqrt{3}}{2} = -5\sqrt{3} \]

Step 3: Verify the relationship between the zeroes and the coefficients:
Let the zeroes of the polynomial be \( \alpha = \sqrt{3} \) and \( \beta = -5\sqrt{3} \). According to Vieta's formulas, the sum and product of the zeroes of the quadratic equation \( at^2 + bt + c = 0 \) are related to the coefficients as follows:
- The sum of the zeroes \( \alpha + \beta = -\frac{b}{a} \) - The product of the zeroes \( \alpha \beta = \frac{c}{a} \) For our equation, \( a = 1 \), \( b = 4\sqrt{3} \), and \( c = -15 \), so we have:
- Sum of the zeroes: \( \alpha + \beta = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} \)
According to Vieta's formula: \( -\frac{b}{a} = -\frac{4\sqrt{3}}{1} = -4\sqrt{3} \), which is correct.
- Product of the zeroes: \( \alpha \beta = \sqrt{3} \times (-5\sqrt{3}) = -5 \times 3 = -15 \)
According to Vieta's formula: \( \frac{c}{a} = \frac{-15}{1} = -15 \), which is also correct.

Conclusion:
The zeroes of the polynomial \( f(t) = t^2 + 4\sqrt{3}t - 15 \) are \( \sqrt{3} \) and \( -5\sqrt{3} \). The sum and product of the zeroes satisfy the relationship with the coefficients as predicted by Vieta's formulas.