Respuesta :
You get the hypothenuse by doing √a²+b² = c
So....
[tex] \sqrt{x^2+[3(x+1)]^2} = 25[/tex]
Square both terms
x² + (3x+3)² = 625
x² + 9x² + 18x + 9 - 625 = 0
10x² + 18x - 616 = 0
x₁,₂ = (-b±√Δ)/2a
Δ = b²-4ac
You call a 10, b 18 and c -616
x1,2 = (-18±√18²-4*10*-616)/2*10
x1,2 = (-18±√324+24640)/20
x1,2 = (-18±√24964)/20
x1,2 = (-18±158)/20
x1 = -18+158/20 = 140/20 = 7
x2 = (-18-158)/20 = -176/20 = -44/5
Pick the first solution
So one leg is 7 and the other is 3(7+1) = 3(8) = 24
Let's verify √24²+7² = √576+49 = √625 = 25
So....
[tex] \sqrt{x^2+[3(x+1)]^2} = 25[/tex]
Square both terms
x² + (3x+3)² = 625
x² + 9x² + 18x + 9 - 625 = 0
10x² + 18x - 616 = 0
x₁,₂ = (-b±√Δ)/2a
Δ = b²-4ac
You call a 10, b 18 and c -616
x1,2 = (-18±√18²-4*10*-616)/2*10
x1,2 = (-18±√324+24640)/20
x1,2 = (-18±√24964)/20
x1,2 = (-18±158)/20
x1 = -18+158/20 = 140/20 = 7
x2 = (-18-158)/20 = -176/20 = -44/5
Pick the first solution
So one leg is 7 and the other is 3(7+1) = 3(8) = 24
Let's verify √24²+7² = √576+49 = √625 = 25
i am not sure about the answer but i guess the answer is 7 and 24
