Question

Can you help me with my math questions

Answer 1

4 part 1: Error was in the second step (2(x)(5)

part 2: Error was in the second step (4c)² - 3³ ≠ (4c - 3)²

5. part 1: 56x² + 60x + 16

part 2: The values of x in the 2 expressions are not the same.

Explanation:

Question 4 part 1

x² - 10x + 25

the factors are -5 and -5

x² - 5x - 5x + 25 = x(x-5) - 5(x-5)

foctoring out (x-5)

= (x-5)(x-5) = (x - 5)²

Question 4 part 2

16c² - 9 = (4c)² - 3²

This is a different of 2 squares

(4c)² - 3² = (4c - 3)(4c + 3)

question 5

part 1

Subtracting the two quadratic expressions

81x² + 90x + 25 - (25x² + 30x + 9)

= 81x² + 90x + 25 - 25x² - 30x - 9

= 56x² + 60x + 16

part 2 solving for x

81x² + 90x + 25 = 1600

81x² + 90x = 1600-25

9x² + 10x = 175

x² + 1.111x + 0.556² = 175 + 0.3086

(x + 0.556)² = 175.3086

x = ±13.2404 - 0.556

= 12.6844 or -13.7964

25x² + 30x + 9 = 484

25x² + 30x = 484 - 9

x² + 1.2x + 0.6² = 19+0.36

(x + 0.6)² = 19.36

x = ±4.4 - 0.6

= 3.8 or -5

Answer 2

Question 4. (x-5)² and (4c-3)(4c+3)

Question 5. Expressions for the areas are 8(7x-1)(x-2) and -8(7x-1)(x-2).

Question 6. Buyer is not right as the values of x are not same.

Explanation:

Question 4). x²- 10x+25 = x²-2×5x+5²

= (x-5)² since (a-b)² = a²+b²-2ab

16c²-9 = (4c)²- 3²

= (4c-3)(4c+3) Since a²-b²=(a-b)(a+b)

Question 5). Area of one square painting = 81x²-90x+25 =(9x-5)²

Area of second square painting = 25x²+30x+9 =(5x+3)²

Now the difference between the areas of these two paintings

= 81x²-90x+25-(25x²+30x+9) = (9x-5)²-(5x+3)²

= (9x-5+5x+3)(9x-5-5x-3) = (14x-2)(4x-8) = 8(7x-1)(x-2)

Since we have to calculate the difference of the areas of two paintings then it may also be written as 25x²+30x+9-(81x²-90x+25)= -8(7x-1)(x-2)

Question.6 The areas are given as 1600 in² and 484 in² then

(9x-5)²= 1600 ⇒ (9x-5) =√1600

9x-5 =40 ⇒ 9x = 40+5 =45

Or x=45/9 =5

and for other painting (5x+3)²=484

5x+3=√484 = 22

5x =22-3

5x =19 ⇒ x=19/5

So the values of x are not same. So buyer is not right.