Here's the question you clicked on:
lawls
If the sides of a right triangle have lengths of x-7, x an x+1, then x= (a) (1, 7) (b) (7, -1) (c) (7, 4) (d) (7, 12) (e) (4, 12)
first tell me which is the longest side
x-7,x,x+1....... what do u thik have the biggest value
Hint: Work with the dearest Pythagorean Theorem.
But beofre using Pythagorean Theorem.. u must find the longest side
Obviously, x + 1 would be the largest side and therefore the hypotenuse because of the + 1 while x is neutral and -7 is negative. THerefore, now use Pythagorean THeorem. \[c^2 = a^2 + b^2\]\[(x + 1)^2 = x^2 + (x - 7)^2\]\[x^2 + 2x + 1 = x^2 + x^2 - 14x + 49\]\[x^2 + 2x + 1 = 2x^2 - 14x + 49\]\[0 = x^2 - 16x + 48\]Can you figure out how to solve the4 quadratic from here?
so answer choice (e) is right?
You, sir, are totally correct!
Kinda ironic since 12 is the only true solution, but whatever works ^_^