If \(x = \frac{{1\; + \;\sqrt 3 }}{2}\) and y = x 3, then y satisfies which one of the following equations?
8y 2– 20y – 1 = 0
Solving the Equation for y given x
The problem asks us to find which equation is satisfied by \(y\), where \(y\) is defined as the cube of \(x\), and \(x\) is given as a specific expression involving a square root.
We are given:
- \(x = \frac{{1\; + \;\sqrt 3 }}{2}\)
- \(y = x^3\)
To find the equation satisfied by \(y\), we first need to calculate the value of \(y\) by cubing \(x\). Then we can substitute this value into the given options or derive the polynomial equation directly from the expression for \(y\).
Step 1: Calculate \(x^2\)
Let's first find \(x^2\), as it will be useful for calculating \(x^3\):
\(x^2 = \left(\frac{{1\; + \;\sqrt 3 }}{2}\right)^2\)
Using the formula \((a+b)^2 = a^2 + 2ab + b^2\):
\(x^2 = \frac{(1)^2 + 2(1)(\sqrt 3) + (\sqrt 3)^2}{2^2}\)
\(x^2 = \frac{1 + 2\sqrt 3 + 3}{4}\)
\(x^2 = \frac{4 + 2\sqrt 3}{4}\)
We can simplify this by dividing the numerator and denominator by 2:
\(x^2 = \frac{2(2 + \sqrt 3)}{4} = \frac{2 + \sqrt 3}{2}\)
Step 2: Calculate \(y = x^3\)
Now we calculate \(y = x^3\) using \(x\) and \(x^2\):
\(y = x^3 = x \cdot x^2\)
\(y = \left(\frac{{1\; + \;\sqrt 3 }}{2}\right) \left(\frac{{2\; + \;\sqrt 3 }}{2}\right)\)
\(y = \frac{(1 + \sqrt 3)(2 + \sqrt 3)}{4}\)
Expand the numerator:
\(y = \frac{1(2) + 1(\sqrt 3) + \sqrt 3(2) + \sqrt 3(\sqrt 3)}{4}\)
\(y = \frac{2 + \sqrt 3 + 2\sqrt 3 + 3}{4}\)
\(y = \frac{5 + 3\sqrt 3}{4}\)
So, the value of \(y\) is \(\frac{5 + 3\sqrt 3}{4}\).
Step 3: Derive the Polynomial Equation for y
We have \(y = \frac{5 + 3\sqrt 3}{4}\). To find a polynomial equation that \(y\) satisfies, we can isolate the square root term and square both sides to eliminate it:
\(4y = 5 + 3\sqrt 3\)
Subtract 5 from both sides:
\(4y - 5 = 3\sqrt 3\)
Now, square both sides of the equation:
\((4y - 5)^2 = (3\sqrt 3)^2\)
Using the formula \((a-b)^2 = a^2 - 2ab + b^2\) on the left side and \((ab)^2 = a^2 b^2\) on the right side:
\((4y)^2 - 2(4y)(5) + (-5)^2 = 3^2 (\sqrt 3)^2\)
\(16y^2 - 40y + 25 = 9 \times 3\)
\(16y^2 - 40y + 25 = 27\)
Subtract 27 from both sides to set the equation to 0:
\(16y^2 - 40y + 25 - 27 = 0\)
\(16y^2 - 40y - 2 = 0\)
We can simplify this quadratic equation by dividing the entire equation by the common factor, 2:
\(\frac{16y^2}{2} - \frac{40y}{2} - \frac{2}{2} = \frac{0}{2}\)
\(8y^2 - 20y - 1 = 0\)
This is a quadratic equation in \(y\). We can now compare this derived equation with the given options.
Step 4: Compare with Options
Let's list the derived equation and the given options:
- Derived Equation: \(8y^2 - 20y - 1 = 0\)
Options:
- \(8y^2 - 20y - 1 = 0\)
- \(8y^2 + 20y - 1 = 0\)
- \(8y^2 + 20y + 1 = 0\)
- \(8y^2 - 20y + 1 = 0\)
Comparing our derived equation \(8y^2 - 20y - 1 = 0\) with the options, we see that it matches the first option.
Thus, \(y\) satisfies the equation \(8y^2 - 20y - 1 = 0\).
Revision Table: Key Steps Summarized
| Step | Description | Calculation |
|---|---|---|
| 1 | Calculate \(x^2\) | \(x^2 = \left(\frac{1 + \sqrt 3}{2}\right)^2 = \frac{2 + \sqrt 3}{2}\) |
| 2 | Calculate \(y = x^3\) | \(y = x \cdot x^2 = \left(\frac{1 + \sqrt 3}{2}\right)\left(\frac{2 + \sqrt 3}{2}\right) = \frac{5 + 3\sqrt 3}{4}\) |
| 3 | Isolate radical in \(y\) expression | \(4y - 5 = 3\sqrt 3\) |
| 4 | Square both sides | \((4y - 5)^2 = (3\sqrt 3)^2 \implies 16y^2 - 40y + 25 = 27\) |
| 5 | Formulate polynomial equation | \(16y^2 - 40y - 2 = 0\) |
| 6 | Simplify equation | \(8y^2 - 20y - 1 = 0\) |
Additional Information: Finding Polynomials Satisfied by Algebraic Numbers
The number \(x = \frac{1 + \sqrt 3}{2}\) is an example of an algebraic number, which is a root of a non-zero polynomial equation with integer coefficients. To find a polynomial equation satisfied by a number involving radicals, a common technique is to isolate the radical term and square both sides repeatedly until all radicals are eliminated. In this problem, \(y = x^3\) is also an algebraic number, and we found the quadratic polynomial \(8y^2 - 20y - 1 = 0\) that it satisfies.
Consider a simple case: Find the equation for \(z = 1 + \sqrt{2}\). \begin{align*} z - 1 &= \sqrt{2} \\ (z - 1)^2 &= (\sqrt{2})^2 \\ z^2 - 2z + 1 &= 2 \\ z^2 - 2z - 1 &= 0 \end{align*} So, \(z\) satisfies the equation \(z^2 - 2z - 1 = 0\). This method is applicable to find the minimum polynomial or other polynomial equations satisfied by expressions involving square roots, cube roots, etc.
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