If 5 and 12 are the two smallest values in a Pythagorean triple, what is the largest value, c, in the triple?
2 answers:
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Answer:
(x - 5)(x + 8)
Step-by-step explanation:
Your equation is: x² + 3x - 40
1. Factor the quadratic
(a) Find two numbers that multiply to give -40 and add to give 3.
List the positive factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
One of the numbers must be negative. Start with the numbers near the middle of the list.
By trial and error, you will find that +8 and -5 work:
-5 × 8 = -40 and -5+8 = 3
(b) Rewrite 3x as -5x + 8x
x² - 5x + 8x - 40
(c) Factor the first two and the last two terms
x(x - 5) + 8(x - 5)
(d) Separate the common factor
x² + 3x - 40 = (x - 5)(x + 8)
2. Using Desmos
You can plot the equation with Desmos.
You get a parabola with x-intercepts at x = -8 and x = 5.
These are the solutions to
x + 8 =0 and x - 5 = 0, so
x² + 3x - 40 = (x+ 8)(x - 5)
