Answer :
Let's find the zeroes of each quadratic polynomial and verify the relationship between the zeroes and the coefficients.
### (i) [tex]\( x^2 - 2x - 8 \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( x^2 - 2x - 8 = 0 \)[/tex]. To find the zeroes, we solve:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -8 \)[/tex].
Solving, we get:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2} \][/tex]
So, the zeroes are:
[tex]\[ x = \frac{2 + 6}{2} = 4 \quad \text{and} \quad x = \frac{2 - 6}{2} = -2 \][/tex]
Thus, the zeroes are [tex]\( x = 4 \)[/tex] and [tex]\( x = -2 \)[/tex].
2. Verify the relationship between the zeroes and coefficients:
- The sum of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) of [tex]\( ax^2 + bx + c = 0 \)[/tex] should be [tex]\( -\frac{b}{a} \)[/tex]:
[tex]\[ \alpha + \beta = - \frac{b}{a} = -\frac{-2}{1} = 2 \][/tex]
Here, the sum is [tex]\( 4 + (-2) = 2 \)[/tex], so it matches.
- The product of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( \frac{c}{a} \)[/tex]:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{-8}{1} = -8 \][/tex]
The product is [tex]\( 4 \times (-2) = -8 \)[/tex], so it matches.
So, for [tex]\( x^2 - 2x - 8 \)[/tex]:
- Zeroes: [tex]\( 4 \)[/tex] and [tex]\( -2 \)[/tex]
- Sum of zeroes: [tex]\( 2 \)[/tex]
- Product of zeroes: [tex]\( -8 \)[/tex]
### (ii) [tex]\( 4s^2r^4 + 4s + 16 \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( 4s^2r^4 + 4s + 16 = 0 \)[/tex].
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 4r^4 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 16 \)[/tex].
Solving, we get:
[tex]\[ s = \frac{-4 \pm \sqrt{16 - 256r^4}}{8r^4} = \frac{-4 \pm \sqrt{4(1 - 16r^4)}}{8r^4} = \frac{-4 \pm 2\sqrt{1 - 16r^4}}{8r^4} = \frac{-2 \pm \sqrt{1 - 16r^4}}{4r^4} \][/tex]
So, the zeroes are:
[tex]\[ s = \frac{-2 + \sqrt{1 - 16r^4}}{4r^4} \quad \text{and} \quad s = \frac{-2 - \sqrt{1 - 16r^4}}{4r^4} \][/tex]
Thus, the zeroes are:
[tex]\[ \left( \frac{-\sqrt{1 - 16r^4} - 1}{2r^4}, \frac{\sqrt{1 - 16r^4} - 1}{2r^4} \right) \][/tex]
### (iii) [tex]\( 6x^2 - 3 - 7x \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( 6x^2 - 3 - 7x = 0 \)[/tex]. Rearranging it to standard form, we get [tex]\( 6x^2 - 7x - 3 = 0 \)[/tex].
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -3 \)[/tex].
Solving, we get:
[tex]\[ x = \frac{7 \pm \sqrt{49 + 72}}{12} = \frac{7 \pm \sqrt{121}}{12} = \frac{7 \pm 11}{12} \][/tex]
So, the zeroes are:
[tex]\[ x = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2} \quad \text{and} \quad x = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3} \][/tex]
Thus, the zeroes are [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( x = -\frac{1}{3} \)[/tex].
2. Verify the relationship between the zeroes and coefficients:
- The sum of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( -\frac{b}{a} \)[/tex]:
[tex]\[ \alpha + \beta = - \frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \][/tex]
Here, the sum is [tex]\( \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \)[/tex], so it matches.
- The product of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( \frac{c}{a} \)[/tex]:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \][/tex]
The product is [tex]\( \left(\frac{3}{2}\right) \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \)[/tex], so it matches.
So, for [tex]\( 6x^2 - 3 - 7x \)[/tex]:
- Zeroes: [tex]\( \frac{3}{2} \)[/tex] and [tex]\( -\frac{1}{3} \)[/tex]
- Sum of zeroes: [tex]\( \frac{7}{6} \)[/tex]
- Product of zeroes: [tex]\( -\frac{1}{2} \)[/tex]
In summary:
1. For [tex]\( x^2 - 2x - 8 \)[/tex]:
- Zeroes: [tex]\(4\)[/tex] and [tex]\(-2\)[/tex]
- Sum of zeroes: [tex]\(2\)[/tex]
- Product of zeroes: [tex]\(-8\)[/tex]
2. For [tex]\( 4s^2r^4 + 4s + 16 \)[/tex]:
- Zeroes: [tex]\(\left(\frac{-\sqrt{1 - 16r^4} - 1}{2r^4}, \frac{\sqrt{1 - 16r^4} - 1}{2r^4}\right)\)[/tex]
3. For [tex]\( 6x^2 - 3 - 7x \)[/tex]:
- Zeroes: [tex]\(\frac{3}{2}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex]
- Sum of zeroes: [tex]\(\frac{7}{6}\)[/tex]
- Product of zeroes: [tex]\(-\frac{1}{2}\)[/tex]
### (i) [tex]\( x^2 - 2x - 8 \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( x^2 - 2x - 8 = 0 \)[/tex]. To find the zeroes, we solve:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -8 \)[/tex].
Solving, we get:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2} \][/tex]
So, the zeroes are:
[tex]\[ x = \frac{2 + 6}{2} = 4 \quad \text{and} \quad x = \frac{2 - 6}{2} = -2 \][/tex]
Thus, the zeroes are [tex]\( x = 4 \)[/tex] and [tex]\( x = -2 \)[/tex].
2. Verify the relationship between the zeroes and coefficients:
- The sum of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) of [tex]\( ax^2 + bx + c = 0 \)[/tex] should be [tex]\( -\frac{b}{a} \)[/tex]:
[tex]\[ \alpha + \beta = - \frac{b}{a} = -\frac{-2}{1} = 2 \][/tex]
Here, the sum is [tex]\( 4 + (-2) = 2 \)[/tex], so it matches.
- The product of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( \frac{c}{a} \)[/tex]:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{-8}{1} = -8 \][/tex]
The product is [tex]\( 4 \times (-2) = -8 \)[/tex], so it matches.
So, for [tex]\( x^2 - 2x - 8 \)[/tex]:
- Zeroes: [tex]\( 4 \)[/tex] and [tex]\( -2 \)[/tex]
- Sum of zeroes: [tex]\( 2 \)[/tex]
- Product of zeroes: [tex]\( -8 \)[/tex]
### (ii) [tex]\( 4s^2r^4 + 4s + 16 \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( 4s^2r^4 + 4s + 16 = 0 \)[/tex].
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 4r^4 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 16 \)[/tex].
Solving, we get:
[tex]\[ s = \frac{-4 \pm \sqrt{16 - 256r^4}}{8r^4} = \frac{-4 \pm \sqrt{4(1 - 16r^4)}}{8r^4} = \frac{-4 \pm 2\sqrt{1 - 16r^4}}{8r^4} = \frac{-2 \pm \sqrt{1 - 16r^4}}{4r^4} \][/tex]
So, the zeroes are:
[tex]\[ s = \frac{-2 + \sqrt{1 - 16r^4}}{4r^4} \quad \text{and} \quad s = \frac{-2 - \sqrt{1 - 16r^4}}{4r^4} \][/tex]
Thus, the zeroes are:
[tex]\[ \left( \frac{-\sqrt{1 - 16r^4} - 1}{2r^4}, \frac{\sqrt{1 - 16r^4} - 1}{2r^4} \right) \][/tex]
### (iii) [tex]\( 6x^2 - 3 - 7x \)[/tex]
1. Find the zeroes:
The quadratic equation is [tex]\( 6x^2 - 3 - 7x = 0 \)[/tex]. Rearranging it to standard form, we get [tex]\( 6x^2 - 7x - 3 = 0 \)[/tex].
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -3 \)[/tex].
Solving, we get:
[tex]\[ x = \frac{7 \pm \sqrt{49 + 72}}{12} = \frac{7 \pm \sqrt{121}}{12} = \frac{7 \pm 11}{12} \][/tex]
So, the zeroes are:
[tex]\[ x = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2} \quad \text{and} \quad x = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3} \][/tex]
Thus, the zeroes are [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( x = -\frac{1}{3} \)[/tex].
2. Verify the relationship between the zeroes and coefficients:
- The sum of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( -\frac{b}{a} \)[/tex]:
[tex]\[ \alpha + \beta = - \frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \][/tex]
Here, the sum is [tex]\( \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \)[/tex], so it matches.
- The product of the zeroes ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) should be [tex]\( \frac{c}{a} \)[/tex]:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \][/tex]
The product is [tex]\( \left(\frac{3}{2}\right) \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \)[/tex], so it matches.
So, for [tex]\( 6x^2 - 3 - 7x \)[/tex]:
- Zeroes: [tex]\( \frac{3}{2} \)[/tex] and [tex]\( -\frac{1}{3} \)[/tex]
- Sum of zeroes: [tex]\( \frac{7}{6} \)[/tex]
- Product of zeroes: [tex]\( -\frac{1}{2} \)[/tex]
In summary:
1. For [tex]\( x^2 - 2x - 8 \)[/tex]:
- Zeroes: [tex]\(4\)[/tex] and [tex]\(-2\)[/tex]
- Sum of zeroes: [tex]\(2\)[/tex]
- Product of zeroes: [tex]\(-8\)[/tex]
2. For [tex]\( 4s^2r^4 + 4s + 16 \)[/tex]:
- Zeroes: [tex]\(\left(\frac{-\sqrt{1 - 16r^4} - 1}{2r^4}, \frac{\sqrt{1 - 16r^4} - 1}{2r^4}\right)\)[/tex]
3. For [tex]\( 6x^2 - 3 - 7x \)[/tex]:
- Zeroes: [tex]\(\frac{3}{2}\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex]
- Sum of zeroes: [tex]\(\frac{7}{6}\)[/tex]
- Product of zeroes: [tex]\(-\frac{1}{2}\)[/tex]
