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Number 2.

Let x be the sides of the equilateral triangle that shortened by 2, 3 and 4 units and a right-triangle is formed.

Let a be the smallest, b a second to the largest, and c is the largest then.

a = x - 4

b = x - 3

c = x - 2

Use Pythagorean Theorem

c² = a² + b²

(x - 2 )² = (x - 4)² + (x - 3)²

x² - 4x + 4 = (x² - 8x + 16) + (x² - 6x + 9)

x² - 4x + 4 = 2x² - 14x + 25

x² - 10x + 21 = 0

(x - 3)(x - 7) = 0

Solve for the value of x

x - 3 = 0

x = 3

&&

x - 7 = 0

x = 7

SS: { 3, 7}

Try to plugged-in the solution sets to the expression given above if it's satisfy.

For x = 3

a = 3 - 4

a = -1, here we can observe that there is no such dimension that has a negative value then we can conclude that the only possible value for the side of equilateral triangle is 7

Answer for 2:

Each sides of an equilateral triangle is 7.

Number 3.

x - √x = 6 >> expression

√x = x - 6 >> square both sides

(√x)² = (x - 6)²

x = x² - 12x + 36 >> let the expression = 0

x² - 13x + 36 = 0 >> factor it.

(x - 4)(x - 9) = 0 >> solve for x.

x - 4 = 0

x = 4

&&

x - 9 = 0

x = 9 then the solution set is

SS: {4, 9}

Plugged the two solutions to the original expression and find if it fits the expression true.

For x = 4

4 - √4 = 6

4 - 2 = 6

2 ≠ 6 ❌

For x = 9

9 - √9 = 6

9 - 3 = 6

6 = 6 ✅

Answer for 3:

The number that has a difference of 6 between its own and its square-root is 9

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