Answer :
To solve the quadratic polynomials and verify the relationships between the zeroes and their coefficients, follow these steps:
### Part (i): [tex]\( 8t^2 + 2t - 15 \)[/tex]
1. Finding the Zeroes:
The quadratic equation is given by [tex]\( 8t^2 + 2t - 15 = 0 \)[/tex]. Solving this equation for [tex]\( t \)[/tex], we find the zeroes to be:
[tex]\[ t_1 = -\frac{3}{2}, \quad t_2 = \frac{5}{4} \][/tex]
2. Sum and Product of the Zeroes:
- Sum of the Zeroes:
[tex]\[ t_1 + t_2 = -\frac{3}{2} + \frac{5}{4} = -\frac{6}{4} + \frac{5}{4} = -\frac{1}{4} \][/tex]
- Product of the Zeroes:
[tex]\[ t_1 \cdot t_2 = \left(-\frac{3}{2}\right) \cdot \left(\frac{5}{4}\right) = -\frac{15}{8} \][/tex]
3. Verification with Coefficients:
For a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], the sum of the zeroes is given by [tex]\(-\frac{b}{a}\)[/tex] and the product is given by [tex]\(\frac{c}{a}\)[/tex].
- Sum Relation:
[tex]\[ -\frac{b}{a} = -\frac{2}{8} = -\frac{1}{4} \][/tex]
We see that [tex]\(-\frac{1}{4}\)[/tex] matches the calculated sum.
- Product Relation:
[tex]\[ \frac{c}{a} = \frac{-15}{8} = -\frac{15}{8} \][/tex]
We see that [tex]\(-\frac{15}{8}\)[/tex] matches the calculated product.
### Part (ii): [tex]\( 4\sqrt{3} x^2 + 5x - 2 \sqrt{3} \)[/tex]
1. Finding the Zeroes:
The quadratic equation is given by [tex]\( 4\sqrt{3} x^2 + 5x - 2 \sqrt{3} = 0 \)[/tex]. Solving this equation for [tex]\( x \)[/tex], we find the zeroes to be:
[tex]\[ x_1 = -\frac{2\sqrt{3}}{3}, \quad x_2 = \frac{\sqrt{3}}{4} \][/tex]
2. Sum and Product of the Zeroes:
- Sum of the Zeroes:
[tex]\[ x_1 + x_2 = -\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{4} = -\frac{8\sqrt{3}}{12} + \frac{3\sqrt{3}}{12} = -\frac{5\sqrt{3}}{12} \][/tex]
- Product of the Zeroes:
[tex]\[ x_1 \cdot x_2 = \left(-\frac{2\sqrt{3}}{3}\right) \cdot \left(\frac{\sqrt{3}}{4}\right) = -\frac{2 \cdot 3}{12} = -\frac{1}{2} \][/tex]
3. Verification with Coefficients:
For a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum of the zeroes is given by [tex]\(-\frac{b}{a}\)[/tex] and the product is given by [tex]\(\frac{c}{a}\)[/tex].
- Sum Relation:
[tex]\[ -\frac{b}{a} = -\frac{5}{4\sqrt{3}} = -\frac{5}{4} \cdot \frac{1}{\sqrt{3}} = -\frac{5\sqrt{3}}{12} \][/tex]
We see that [tex]\(-\frac{5\sqrt{3}}{12}\)[/tex] matches the calculated sum.
- Product Relation:
[tex]\[ \frac{c}{a} = \frac{-2\sqrt{3}}{4\sqrt{3}} = -\frac{2\sqrt{3}}{4\sqrt{3}} = -\frac{2}{4} = -\frac{1}{2} \][/tex]
We see that [tex]\(-\frac{1}{2}\)[/tex] matches the calculated product.
Thus, we have found the zeroes of both quadratic polynomials and verified the relations between their zeroes and coefficients.
### Part (i): [tex]\( 8t^2 + 2t - 15 \)[/tex]
1. Finding the Zeroes:
The quadratic equation is given by [tex]\( 8t^2 + 2t - 15 = 0 \)[/tex]. Solving this equation for [tex]\( t \)[/tex], we find the zeroes to be:
[tex]\[ t_1 = -\frac{3}{2}, \quad t_2 = \frac{5}{4} \][/tex]
2. Sum and Product of the Zeroes:
- Sum of the Zeroes:
[tex]\[ t_1 + t_2 = -\frac{3}{2} + \frac{5}{4} = -\frac{6}{4} + \frac{5}{4} = -\frac{1}{4} \][/tex]
- Product of the Zeroes:
[tex]\[ t_1 \cdot t_2 = \left(-\frac{3}{2}\right) \cdot \left(\frac{5}{4}\right) = -\frac{15}{8} \][/tex]
3. Verification with Coefficients:
For a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], the sum of the zeroes is given by [tex]\(-\frac{b}{a}\)[/tex] and the product is given by [tex]\(\frac{c}{a}\)[/tex].
- Sum Relation:
[tex]\[ -\frac{b}{a} = -\frac{2}{8} = -\frac{1}{4} \][/tex]
We see that [tex]\(-\frac{1}{4}\)[/tex] matches the calculated sum.
- Product Relation:
[tex]\[ \frac{c}{a} = \frac{-15}{8} = -\frac{15}{8} \][/tex]
We see that [tex]\(-\frac{15}{8}\)[/tex] matches the calculated product.
### Part (ii): [tex]\( 4\sqrt{3} x^2 + 5x - 2 \sqrt{3} \)[/tex]
1. Finding the Zeroes:
The quadratic equation is given by [tex]\( 4\sqrt{3} x^2 + 5x - 2 \sqrt{3} = 0 \)[/tex]. Solving this equation for [tex]\( x \)[/tex], we find the zeroes to be:
[tex]\[ x_1 = -\frac{2\sqrt{3}}{3}, \quad x_2 = \frac{\sqrt{3}}{4} \][/tex]
2. Sum and Product of the Zeroes:
- Sum of the Zeroes:
[tex]\[ x_1 + x_2 = -\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{4} = -\frac{8\sqrt{3}}{12} + \frac{3\sqrt{3}}{12} = -\frac{5\sqrt{3}}{12} \][/tex]
- Product of the Zeroes:
[tex]\[ x_1 \cdot x_2 = \left(-\frac{2\sqrt{3}}{3}\right) \cdot \left(\frac{\sqrt{3}}{4}\right) = -\frac{2 \cdot 3}{12} = -\frac{1}{2} \][/tex]
3. Verification with Coefficients:
For a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum of the zeroes is given by [tex]\(-\frac{b}{a}\)[/tex] and the product is given by [tex]\(\frac{c}{a}\)[/tex].
- Sum Relation:
[tex]\[ -\frac{b}{a} = -\frac{5}{4\sqrt{3}} = -\frac{5}{4} \cdot \frac{1}{\sqrt{3}} = -\frac{5\sqrt{3}}{12} \][/tex]
We see that [tex]\(-\frac{5\sqrt{3}}{12}\)[/tex] matches the calculated sum.
- Product Relation:
[tex]\[ \frac{c}{a} = \frac{-2\sqrt{3}}{4\sqrt{3}} = -\frac{2\sqrt{3}}{4\sqrt{3}} = -\frac{2}{4} = -\frac{1}{2} \][/tex]
We see that [tex]\(-\frac{1}{2}\)[/tex] matches the calculated product.
Thus, we have found the zeroes of both quadratic polynomials and verified the relations between their zeroes and coefficients.
