SOLUTION: Ok the question is they gave me a diagram of a right angle triangle ABC. X-7 is the adjacent,x is the opposite and x+1 is my hypotenuse.The question is show that x^2-16x+48=0.i tri

Question 892105: Ok the question is they gave me a diagram of a right angle triangle ABC. X-7 is the adjacent,x is the opposite and x+1 is my hypotenuse.The question is show that x^2-16x+48=0.i tried Pythagoras theorem but failed.how to derive that ans x^2-16x+48=0.HELP TMR IS MY EXAM URGENT HELP PLEASE
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you use the pythagorean theorem.

that states that a^2 + b^2 = c^2

c is the hypotenuse.
a is one of the legs
b is the other of the legs.

since there's no angle involved, it doesn't matter which is the adjacent side or the opposite side.

that only comes into play when you are looking for a trig function related to that triangle.

in your problem, you can make:

a = x-7
b = x
c = x+1

the only restriction is that c had to be equal to x+1 because c is the hypotenuse of the right triangle.

a^2 + b^2 = c^2 becomes (x-7)^2 + x^2 = (x+1)^2

multiply these factors out and you get:

x^2 - 14x + 49 + x^2 = x^2 + 2x + 1

subtract (x^2 + 2x + 1) from both sides of the equation and you are left with:

x^2 - 14x + 49 + x^2 - (x^2 + 2x + 1) = 0

simplify this to get:

x^2 - 14x + 49 + x^2 - x^2 - 2x - 1 = 0

combine like terms to get:

x^2 - 16x + 48 = 0

i believe that's the equation you are looking for.

you had the right idea using the pythagorean theorem but you failed somewhere in the execution.

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