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Exercise 5

We are given $C = (5x - 3)^2 - (2x - 1)(5x - 3)$. Let's solve step-by-step:

1. Develop, reduce, and organize $C$:

• Expand each term: $(5x - 3)^2 = 25x^2 - 30x + 9$ $(2x - 1)(5x - 3) = 10x^2 - 11x + 3$

• Subtract: $C = (25x^2 - 30x + 9) - (10x^2 - 11x + 3)$ $C = 25x^2 - 30x + 9 - 10x^2 + 11x - 3$ $C = 15x^2 - 19x + 6$

2. Calculate $C$ for $x = -2$:

Substitute $x = -2$ into $C = 15x^2 - 19x + 6$: $C = 15(-2)^2 - 19(-2) + 6$ $C = 15(4) + 38 + 6 = 60 + 38 + 6 = 104$

3. Factorize $C$:

The quadratic $C = 15x^2 - 19x + 6$ factors as: $C = (3x - 2)(5x - 3)$

4. Solve the equation $(5x - 3)(2 - 3x) = 0$:

Set each factor to 0: $5x - 3 = 0 \quad \text{or} \quad 2 - 3x = 0$ $x = \frac{3}{5} \quad \text{or} \quad x = \frac{2}{3}$

Exercise 6

1. Simplify $A, B, C$ without direct multiplication:

• $A = 109 \times 91$: Use the difference of squares: $109 \times 91 = (100 + 9)(100 - 9) = 100^2 - 9^2 = 10000 - 81 = 9919$

• $B = 125^2 - 75^2$: Again, use the difference of squares: $B = (125 - 75)(125 + 75) = 50 \times 200 = 10000$

• $C = 103^2$: Use the formula $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 100$ and $b = 3$: $C = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609$

2. Calculate $x(3x - 1) - 2(4x - 2)$ for $x = -1$:

Substitute $x = -1$: $x(3x - 1) - 2(4x - 2) = (-1)(3(-1) - 1) - 2(4(-1) - 2)$ $= (-1)(-3 - 1) - 2(-4 - 2) = (-1)(-4) - 2(-6)$ $= 4 + 12 = 16$

3. Calculate $(3x - 4)^2 - (2x + 1)^2$ for $x = -2$:

Substitute $x = -2$: $(3x - 4)^2 - (2x + 1)^2 = (3(-2) - 4)^2 - (2(-2) + 1)^2$ $= (-6 - 4)^2 - (-4 + 1)^2 = (-10)^2 - (-3)^2$ $= 100 - 9 = 91$

Exercise 7

1. Express $A_1$ (area of rectangle $ABCD$) in terms of $x$:

The rectangle's dimensions are $2x + 1$ (length) and $3x - 5$ (width): $A_1 = (2x + 1)(3x - 5)$

2. Express $A_2$ (area of polygon $EFGHIJ$) in terms of $x$:

The total area is made by adding the squares and subtracting the rectangle: $A_2 = (3x - 7)^2 - z^2$ where $z = (3x - 7) - (2x + 1) = x - 8$, so $z^2 = (x - 8)^2$: $A_2 = (3x - 7)^2 - (x - 8)^2$

Factorize using the difference of squares: [ A_2 = [(3x - 7) - (x - 8)][(3x - 7) + (x - 8)] ] $A_2 = (2x + 1)(4x - 15)$

3. Find $x$ such that $A_1 = A_2$:

Set $(2x + 1)(3x - 5) = (2x + 1)(4x - 15)$. Divide by $2x + 1$ (valid if $x \neq -\frac{1}{2}$): $3x - 5 = 4x - 15$ $x = 10$

Verify that $x = 10$ satisfies the original expressions and is valid for the geometric conditions.

Let me know if you want further details or explanations!

Here are 5 related questions for further practice:

1. How do you systematically expand and simplify algebraic expressions?
2. Can the area formulas in Exercise 7 be further simplified geometrically?
3. What is the significance of the difference of squares in simplifying calculations?
4. How do you check if solutions satisfy the conditions of a problem?
5. Why is factoring an essential skill in solving quadratic equations?

Tip: Always verify your final solutions by substituting back into the original problem to ensure consistency.