Answer :
Alright! Let's dive into each part of the question step by step:
### Part (a)
#### Given:
[tex]\[ x(x + 3) = p \][/tex]
1. When [tex]\( p = 0 \)[/tex]:
[tex]\[ x(x + 3) = 0 \][/tex]
For the product of two numbers to be zero, at least one of the numbers must be zero. Therefore, the solutions are:
[tex]\[ x = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x = -3 \][/tex]
Thus, the solutions are:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
2. When [tex]\( p = x + 3 \)[/tex]:
[tex]\[ x(x + 3) = x + 3 \][/tex]
First, rearrange the equation:
[tex]\[ x^2 + 3x = x + 3 \][/tex]
Subtract [tex]\( x + 3 \)[/tex] from both sides:
[tex]\[ x^2 + 3x - x - 3 = 0 \][/tex]
[tex]\[ x^2 + 2x - 3 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex].
To solve it, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 4}{2} \][/tex]
So the solutions are:
[tex]\[ x = \frac{2}{2} = 1 \][/tex]
[tex]\[ x = \frac{-6}{2} = -3 \][/tex]
Thus, the solutions are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = -3 \][/tex]
### Part (b)
1. Solve [tex]\( 3 \log_2(x - 2) = 9 \)[/tex]:
First, divide both sides by 3:
[tex]\[ \log_2(x - 2) = 3 \][/tex]
We convert this logarithmic equation to its exponential equivalent:
[tex]\[ x - 2 = 2^3 \][/tex]
[tex]\[ x - 2 = 8 \][/tex]
Add 2 to both sides:
[tex]\[ x = 10 \][/tex]
Thus, the solution is:
[tex]\[ x = 10 \][/tex]
2. Solve [tex]\( x^2 - 6x < 16 \)[/tex]:
First, move 16 to the left side of the inequality:
[tex]\[ x^2 - 6x - 16 < 0 \][/tex]
This is a quadratic inequality, and to solve it, we need to find the roots of the quadratic equation:
[tex]\[ x^2 - 6x - 16 = 0 \][/tex]
We use the quadratic formula again:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -16 \)[/tex]:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{6 \pm 10}{2} \][/tex]
So the solutions for the roots are:
[tex]\[ x = \frac{16}{2} = 8 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
Since the quadratic expression [tex]\( x^2 - 6x - 16 \)[/tex] opens upwards (as the coefficient of [tex]\( x^2 \)[/tex] is positive), the solution to the inequality [tex]\( x^2 - 6x - 16 < 0 \)[/tex] is the interval between the roots:
[tex]\[ -2 < x < 8 \][/tex]
Thus, the solution is:
[tex]\[ x \in (-2, 8) \][/tex]
### Part (c)
Given the quadratic equation:
[tex]\[ x^2 - 6ax + 5 = 0 \][/tex]
1. Determine the roots in terms of [tex]\( a \)[/tex]:
Using the quadratic formula:
[tex]\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Where [tex]\( A = 1 \)[/tex], [tex]\( B = -6a \)[/tex], and [tex]\( C = 5 \)[/tex]:
[tex]\[ x = \frac{6a \pm \sqrt{(6a)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{6a \pm \sqrt{36a^2 - 20}}{2} \][/tex]
[tex]\[ x = \frac{6a \pm \sqrt{36a^2 - 20}}{2} \][/tex]
Thus, the roots in terms of [tex]\( a \)[/tex] are:
[tex]\[ x = \frac{6a + \sqrt{36a^2 - 20}}{2} \][/tex]
[tex]\[ x = \frac{6a - \sqrt{36a^2 - 20}}{2} \][/tex]
2. Show that the product of the roots is independent of [tex]\( a \)[/tex]:
The product of the roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ x_1 \cdot x_2 = \frac{c}{a} \][/tex]
For [tex]\( x^2 - 6ax + 5 = 0 \)[/tex]:
[tex]\[ a = 1 \][/tex]
[tex]\[ c = 5 \][/tex]
Thus, the product of the roots is:
[tex]\[ x_1 \cdot x_2 = \frac{5}{1} = 5 \][/tex]
As we can see, the product of the roots is 5, which is independent of [tex]\( a \)[/tex].
### Part (a)
#### Given:
[tex]\[ x(x + 3) = p \][/tex]
1. When [tex]\( p = 0 \)[/tex]:
[tex]\[ x(x + 3) = 0 \][/tex]
For the product of two numbers to be zero, at least one of the numbers must be zero. Therefore, the solutions are:
[tex]\[ x = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x = -3 \][/tex]
Thus, the solutions are:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
2. When [tex]\( p = x + 3 \)[/tex]:
[tex]\[ x(x + 3) = x + 3 \][/tex]
First, rearrange the equation:
[tex]\[ x^2 + 3x = x + 3 \][/tex]
Subtract [tex]\( x + 3 \)[/tex] from both sides:
[tex]\[ x^2 + 3x - x - 3 = 0 \][/tex]
[tex]\[ x^2 + 2x - 3 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex].
To solve it, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 4}{2} \][/tex]
So the solutions are:
[tex]\[ x = \frac{2}{2} = 1 \][/tex]
[tex]\[ x = \frac{-6}{2} = -3 \][/tex]
Thus, the solutions are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = -3 \][/tex]
### Part (b)
1. Solve [tex]\( 3 \log_2(x - 2) = 9 \)[/tex]:
First, divide both sides by 3:
[tex]\[ \log_2(x - 2) = 3 \][/tex]
We convert this logarithmic equation to its exponential equivalent:
[tex]\[ x - 2 = 2^3 \][/tex]
[tex]\[ x - 2 = 8 \][/tex]
Add 2 to both sides:
[tex]\[ x = 10 \][/tex]
Thus, the solution is:
[tex]\[ x = 10 \][/tex]
2. Solve [tex]\( x^2 - 6x < 16 \)[/tex]:
First, move 16 to the left side of the inequality:
[tex]\[ x^2 - 6x - 16 < 0 \][/tex]
This is a quadratic inequality, and to solve it, we need to find the roots of the quadratic equation:
[tex]\[ x^2 - 6x - 16 = 0 \][/tex]
We use the quadratic formula again:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -16 \)[/tex]:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{6 \pm 10}{2} \][/tex]
So the solutions for the roots are:
[tex]\[ x = \frac{16}{2} = 8 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
Since the quadratic expression [tex]\( x^2 - 6x - 16 \)[/tex] opens upwards (as the coefficient of [tex]\( x^2 \)[/tex] is positive), the solution to the inequality [tex]\( x^2 - 6x - 16 < 0 \)[/tex] is the interval between the roots:
[tex]\[ -2 < x < 8 \][/tex]
Thus, the solution is:
[tex]\[ x \in (-2, 8) \][/tex]
### Part (c)
Given the quadratic equation:
[tex]\[ x^2 - 6ax + 5 = 0 \][/tex]
1. Determine the roots in terms of [tex]\( a \)[/tex]:
Using the quadratic formula:
[tex]\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Where [tex]\( A = 1 \)[/tex], [tex]\( B = -6a \)[/tex], and [tex]\( C = 5 \)[/tex]:
[tex]\[ x = \frac{6a \pm \sqrt{(6a)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{6a \pm \sqrt{36a^2 - 20}}{2} \][/tex]
[tex]\[ x = \frac{6a \pm \sqrt{36a^2 - 20}}{2} \][/tex]
Thus, the roots in terms of [tex]\( a \)[/tex] are:
[tex]\[ x = \frac{6a + \sqrt{36a^2 - 20}}{2} \][/tex]
[tex]\[ x = \frac{6a - \sqrt{36a^2 - 20}}{2} \][/tex]
2. Show that the product of the roots is independent of [tex]\( a \)[/tex]:
The product of the roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ x_1 \cdot x_2 = \frac{c}{a} \][/tex]
For [tex]\( x^2 - 6ax + 5 = 0 \)[/tex]:
[tex]\[ a = 1 \][/tex]
[tex]\[ c = 5 \][/tex]
Thus, the product of the roots is:
[tex]\[ x_1 \cdot x_2 = \frac{5}{1} = 5 \][/tex]
As we can see, the product of the roots is 5, which is independent of [tex]\( a \)[/tex].
