a. Given that p = x + ym\(^3\), find m in terms of p, x, and y
b. Using the method of completing the square method, find the roots of the quadratic equation x\(^2\) - 6x + 7 = 0 to 1 decimal place.
c. The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it. Find the two numbers.
a. p = x + ym\(^3\)
ym\(^3\) = p - x
m\(^3\) = \(\frac{p - x}{\text{y}}\)
\(m = \sqrt[3]{\frac{p - x}{y}}\).
b. x\(^2\) - 6x + 7 = 0
Using the completing the square method
x\(^2\) - 6x = - 7
Add the square of half the coefficient of x to both sides
x\(^2\) - 6x + (-3)\(^2\) = -7 + 9
x\(^2\) - 6x + (-3)\(^2\) = 2
(x - 3)\(^2\) = 2
x - 3 = ± \(\sqrt{2}\)
x = 3 + \(\sqrt{2}\) or 3 - \(\sqrt{2}\)
x = 3 + 1.4142 or 3 - 1.4142
x = 4.1 or 1.6 to 1 decimal place.
c. Let the two consecutive positive odd numbers be \(n\) and \(n + 2\).
The equation for their product is:
\(n(n + 2) = 195\)
Expanding gives:
n\(^2\) + 2n - 195 = 0
Using the quadratic formula:
\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
with \(a = 1\), \(b = 2\), and \(c = -195\):
\(b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-195) = 4 + 780 = 784\)
Now substituting: \(n = \frac{-2 \pm \sqrt{784}}{2} = \frac{-2 \pm 28}{2}\)
This results in: \(n = \frac{26}{2} = 13 \quad (\text{positive solution})\)
Thus, the two numbers are: \(13 \quad \text{and} \quad 15\)
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