Question

Pls help me I’m so stuck :(

Answer 1

B

Explanation:

using Pythagoras' identity in the right triangle

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , then

(x + 1)² + (x + 3)² = (x + 5)² ← expand all factors using FOIL

x² + 2x + 1 + x² + 6x + 9 = x² + 10x + 25 , that is

2x² + 8x + 10 = x² + 10x + 25 ← subtract x² + 10x + 25 from both sides

x² - 2x - 15 = 0 ← in standard form

(x - 5)(x + 3) = 0 ← in factored form

equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 3 = 0 ⇒ x = - 3

however, x > 0 , then x = 5

Answer 2

B. 5

Explanation:

`\sf{C = \sqrt{ A {}^(2) + B {}^(2) }}`

`\sf{x + 5 = \sqrt{ {(x + 3)}^(2) + {(x + 1)}^(2) }}`

`\sf{x + 5 = \sqrt{ {(x {}^(2) + 6x + 9)} + {(x {}^(2) + 2x + 1)}}}`

`\sf{x + 5 = \sqrt{ {(2x {}^(2) + 8x + 10)}}}`

`\sf{(x + 5) {}^(2) = { {2x {}^(2) + 8x + 10}}}`

`\sf{ {x}^(2) + 10x + 25= { {2x {}^(2) + 8x + 10}}}`

`\sf{0= { {2x {}^(2) \red{ { - x}^(2) } + 8x \red{ - 10x} + 10 \red{ - 25}}}}`

`\sf{0= x^(2) } - 2x { - 15}`

`\sf{0= x^(2) } - 2x { - 15}`

`\sf{0= (x - 5)(x + 3)}`

`\sf{x_(1) - 5 = 0 }`

`\sf{x_(1) = 0 + 5 }`

`\sf{x_(1) = \boxed{5 }}`

`\sf{x_(2) + 3 = 0 }`

`\sf{x_(2) = 0 - 3 }`

`\sf{x_(2) = \boxed{- 3 }}`

Option B. 5