Let the width be x.

Length is 8 feet more than width. Length = x + 8

Area = x(x + 8)

width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)

Area = (x + 4)(x +3)

Area remains the same

x(x + 8) = (x+4)(x +3)

x² + 8x = x(x +3) + 4(x +3)

x² + 8x = x² +3x + 4x +12

x² + 8x = x² +7x +12 Eliminate x² from both sides

8x = 7x + 12

8x - 7x = 12

x = 12

Dimensions of original rectangle : x, x + 8

12, 12 +8 = 12, 20

Original rectangle is 20 feet by 12 feet

5 0

3 years ago

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How many real-number solutions does the equation have?

Nataly_w [17]

Answer:

Option b. Two solutions

Step-by-step explanation:

In order to find how many real number solutions the equation has we have to solve it

Given equation: -4x² + 10x + 6 = 0

taking 2 common from the equation

2(-2x² + 5x + 3) = 0

-2x² + 5x + 3 = 0

taking minus sign common from the above equation

2x² - 5x - 3 = 0

We will solve this equation by factorization in such a way that the sum of two factors is equal to -5x and the product is -6x²

2x² - 6x + x - 3 = 0

taking common above

2x(x-3) + 1(x-3) = 0

taking (x-3) common

(2x+1)(x-3) = 0

2x + 1 = 0

2x = -1

x = $\frac{-1}{2}$

x - 3 = 0; x = 3

the solutions are

$x=\frac{-1}{2},3$

Both values are real numbers, therefore correct option is b

7 0

3 years ago

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