Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
