Question I need some help on the steps to find the missing sides of a rt triangle c= 10, a = x-2 and b = x-4.
Answer For a right triangle, we know that a² + b² = c².
Putting in what we're given for a, b, and c gives us (x-2)² + (x-4)² = 10².
Multiplying this out gives x² - 4x + 4 + x² - 8x + 16 = 100.
Combining like terms gives 2x² - 12x - 80 = 0.
The coefficients are all even, so dividing the equation by 2 gives x² - 6x - 40 = 0.
Note that here we have a = 1, b = -6, and c = -40.
The a, b, and c of the original problem are different, but we'll put them on hold.
Think of this as calling a subroutine and giving the subroutine local variables a, b, and c.
They disappear once we get back to the main program.
Using the quadratic formula, which is (-b ± √(b²-4ac))/(2a), we get
(6 ± √(36 + 160))/2 = (6 ± 14)/2 = 3 ± 7.
This gives two choices, only on of which gives positive values.
x = 3+7 = 10, which is c, a = x-2 = 10-2 = 8, and b = x-4 = 10-4 = 6.
On the other choice, it would be like our distance were directional.
x would be 3-7 = -4. The value of c is still 10, but now we have a = x-2 = -4-2 = -6
and b = x-4 = -4-4 = -8. In other words, we have a triangle in quadrant III on a x-y graph
and the first triangle is in quadrant I.
