Answer :
Let's go through each part of the problem step-by-step.
Q1(a)(i): Factorise [tex]\(x^2 - 12x + 27\)[/tex]
To factor the quadratic expression [tex]\(x^2 - 12x + 27\)[/tex], we look for two numbers that multiply to 27 (the constant term) and add to -12 (the coefficient of x).
The numbers -3 and -9 satisfy these conditions because:
- [tex]\((-3) \times (-9) = 27\)[/tex]
- [tex]\((-3) + (-9) = -12\)[/tex]
Therefore, we can factor the expression as:
[tex]\[
x^2 - 12x + 27 = (x - 3)(x - 9)
\][/tex]
Q1(a)(ii): Solve the equation [tex]\(x^2 - 12x + 27 = 0\)[/tex]
We already factored the quadratic:
[tex]\[
(x - 3)(x - 9) = 0
\][/tex]
To solve, set each factor to zero:
1. [tex]\(x - 3 = 0\)[/tex] gives [tex]\(x = 3\)[/tex]
2. [tex]\(x - 9 = 0\)[/tex] gives [tex]\(x = 9\)[/tex]
The solutions to the equation are [tex]\(x = 3\)[/tex] and [tex]\(x = 9\)[/tex].
Q1(b): Factorise [tex]\(y^2 - 100\)[/tex]
The expression [tex]\(y^2 - 100\)[/tex] is a difference of squares. It can be factored using the formula [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex].
Here, [tex]\(y^2 - 100\)[/tex] is:
[tex]\[
y^2 - 10^2 = (y - 10)(y + 10)
\][/tex]
Q2: Solve, by factorising, the equation [tex]\(8x^2 - 30x - 27 = 0\)[/tex]
We need to factor this quadratic expression. The factors are [tex]\((2x - 9)\)[/tex] and [tex]\((4x + 3)\)[/tex].
[tex]\[
(2x - 9)(4x + 3) = 0
\][/tex]
To find the solutions, set each factor to zero:
1. [tex]\(2x - 9 = 0\)[/tex] gives [tex]\(2x = 9\)[/tex], so [tex]\(x = \frac{9}{2}\)[/tex]
2. [tex]\(4x + 3 = 0\)[/tex] gives [tex]\(4x = -3\)[/tex], so [tex]\(x = -\frac{3}{4}\)[/tex]
Thus, the solutions to the equation are [tex]\(x = \frac{9}{2}\)[/tex] and [tex]\(x = -\frac{3}{4}\)[/tex].
I hope this helps clarify the solutions! Let me know if you have any questions.
Q1(a)(i): Factorise [tex]\(x^2 - 12x + 27\)[/tex]
To factor the quadratic expression [tex]\(x^2 - 12x + 27\)[/tex], we look for two numbers that multiply to 27 (the constant term) and add to -12 (the coefficient of x).
The numbers -3 and -9 satisfy these conditions because:
- [tex]\((-3) \times (-9) = 27\)[/tex]
- [tex]\((-3) + (-9) = -12\)[/tex]
Therefore, we can factor the expression as:
[tex]\[
x^2 - 12x + 27 = (x - 3)(x - 9)
\][/tex]
Q1(a)(ii): Solve the equation [tex]\(x^2 - 12x + 27 = 0\)[/tex]
We already factored the quadratic:
[tex]\[
(x - 3)(x - 9) = 0
\][/tex]
To solve, set each factor to zero:
1. [tex]\(x - 3 = 0\)[/tex] gives [tex]\(x = 3\)[/tex]
2. [tex]\(x - 9 = 0\)[/tex] gives [tex]\(x = 9\)[/tex]
The solutions to the equation are [tex]\(x = 3\)[/tex] and [tex]\(x = 9\)[/tex].
Q1(b): Factorise [tex]\(y^2 - 100\)[/tex]
The expression [tex]\(y^2 - 100\)[/tex] is a difference of squares. It can be factored using the formula [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex].
Here, [tex]\(y^2 - 100\)[/tex] is:
[tex]\[
y^2 - 10^2 = (y - 10)(y + 10)
\][/tex]
Q2: Solve, by factorising, the equation [tex]\(8x^2 - 30x - 27 = 0\)[/tex]
We need to factor this quadratic expression. The factors are [tex]\((2x - 9)\)[/tex] and [tex]\((4x + 3)\)[/tex].
[tex]\[
(2x - 9)(4x + 3) = 0
\][/tex]
To find the solutions, set each factor to zero:
1. [tex]\(2x - 9 = 0\)[/tex] gives [tex]\(2x = 9\)[/tex], so [tex]\(x = \frac{9}{2}\)[/tex]
2. [tex]\(4x + 3 = 0\)[/tex] gives [tex]\(4x = -3\)[/tex], so [tex]\(x = -\frac{3}{4}\)[/tex]
Thus, the solutions to the equation are [tex]\(x = \frac{9}{2}\)[/tex] and [tex]\(x = -\frac{3}{4}\)[/tex].
I hope this helps clarify the solutions! Let me know if you have any questions.
