What are the possible number of positive real, negative real, and complex zeros of f(x) = 4x3 + x2 + 10x – 14?
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The length of the shorter leg is 14 inches.
We will set up two equations for this; first, for the perimeter. We will call the legs a and b:
a + b + 50 = 112
Subtract 50 from both sides:
a + b + 50 - 50 = 112 - 50
a + b = 62
We will isolate a, so we can use substitution. Subtract b from both sides:
a + b - b = 62 - b
a = 62 - b
Now we will set up our Pythagorean theorem equation:
a² + b² = 50²
a² + b² = 2500
We will substitute our value for a from the first equation into this one:
(62-b)² + b² = 2500
(62 - b)(62 - b) + b² = 2500
Multiplying the binomials, we have:
62*62 - b*62 - b*62 -b(-b) + b² = 2500
3844 - 62b - 62b + b² + b² = 2500
Combining like terms:
3844 - 124b + 2b² = 2500
To set it equal to 0 and solve, we subtract 2500 from both sides:
3844 - 124b + 2b² - 2500 = 2500 - 2500
1344 - 124b + 2b² = 0
In standard form, we have
2b² - 124b + 1344 = 0
Using the quadratic formula,
The shorter leg is 14 and the longer one is 48.
We will set up two equations for this; first, for the perimeter. We will call the legs a and b:
a + b + 50 = 112
Subtract 50 from both sides:
a + b + 50 - 50 = 112 - 50
a + b = 62
We will isolate a, so we can use substitution. Subtract b from both sides:
a + b - b = 62 - b
a = 62 - b
Now we will set up our Pythagorean theorem equation:
a² + b² = 50²
a² + b² = 2500
We will substitute our value for a from the first equation into this one:
(62-b)² + b² = 2500
(62 - b)(62 - b) + b² = 2500
Multiplying the binomials, we have:
62*62 - b*62 - b*62 -b(-b) + b² = 2500
3844 - 62b - 62b + b² + b² = 2500
Combining like terms:
3844 - 124b + 2b² = 2500
To set it equal to 0 and solve, we subtract 2500 from both sides:
3844 - 124b + 2b² - 2500 = 2500 - 2500
1344 - 124b + 2b² = 0
In standard form, we have
2b² - 124b + 1344 = 0
Using the quadratic formula,
The shorter leg is 14 and the longer one is 48.