If \({\left( {a + \;\frac{1}{a} + 3} \right)^2}\) = 16, where a is a non-zero real number, then find the value of a 2 + \(\frac{1}{{{a^2}}}\) .
47
Understanding the Given Algebraic Equation
The problem asks us to find the value of \(a^2 + \frac{1}{a^2}\) given an equation involving \(a + \frac{1}{a}\) and a constant. The given equation is:
\({\left( {a + \;\frac{1}{a} + 3} \right)^2}\) = 16
We are also told that 'a' is a non-zero real number. This condition is important because it affects the possible values of expressions like \(a + \frac{1}{a}\) and \(a^2 + \frac{1}{a^2}\).
Solving for the Value of \(a + \frac{1}{a}\)
To find the value of \(a^2 + \frac{1}{a^2}\), we first need to determine the possible value(s) of \(a + \frac{1}{a}\). We can do this by taking the square root of both sides of the given equation:
\(\sqrt{{{\left( {a + \;\frac{1}{a} + 3} \right)}^2}}} = \sqrt{16}\)
\(a + \frac{1}{a} + 3 = \pm 4\)
This gives us two possible cases:
- \(a + \frac{1}{a} + 3 = 4\)
- \(a + \frac{1}{a} + 3 = -4\)
Case 1: \(a + \frac{1}{a} + 3 = 4\)
Subtracting 3 from both sides:
\(a + \frac{1}{a} = 4 - 3\)
\(a + \frac{1}{a} = 1\)
Case 2: \(a + \frac{1}{a} + 3 = -4\)
Subtracting 3 from both sides:
\(a + \frac{1}{a} = -4 - 3\)
\(a + \frac{1}{a} = -7\)
Analysing the Possible Values for \(a + \frac{1}{a}\) for Real 'a'
For a real number 'a' (where \(a \ne 0\)), the expression \(a + \frac{1}{a}\) has a specific property. If \(a + \frac{1}{a} = k\), we can multiply by 'a' to get \(a^2 + 1 = ka\), which rearranges to \(a^2 - ka + 1 = 0\). For this quadratic equation to have real roots for 'a', the discriminant must be non-negative. The discriminant is \(D = {(-k)}^2 - 4(1)(1) = k^2 - 4\). Thus, for real 'a', we must have \(k^2 - 4 \ge 0\), which means \(k^2 \ge 4\). This inequality holds true if and only if \(k \ge 2\) or \(k \le -2\).
Let's check our values of \(a + \frac{1}{a}\):
- In Case 1, \(a + \frac{1}{a} = 1\). Here, \(k=1\). Since \(1^2 = 1\), and \(1 < 4\), this value \(a + \frac{1}{a} = 1\) is not possible for any real number 'a'.
- In Case 2, \(a + \frac{1}{a} = -7\). Here, \(k=-7\). Since \({(-7)}^2 = 49\), and \(49 \ge 4\), this value \(a + \frac{1}{a} = -7\) is possible for some real number 'a'.
Therefore, the only valid value for \(a + \frac{1}{a}\) in this problem, given that 'a' is a real number, is -7.
Evaluating \(a^2 + \frac{1}{a^2}\) using Algebraic Identities
We need to find the value of \(a^2 + \frac{1}{a^2}\). We know a standard algebraic identity that relates \({\left( {a + \frac{1}{a}} \right)^2}\) to \(a^2 + \frac{1}{a^2}\):
\({\left( {a + \frac{1}{a}} \right)^2} = a^2 + 2 \cdot a \cdot \frac{1}{a} + \left( \frac{1}{a} \right)^2\)
\({\left( {a + \frac{1}{a}} \right)^2} = a^2 + 2 + \frac{1}{a^2}\)
Rearranging this identity to solve for \(a^2 + \frac{1}{a^2}\):
\(a^2 + \frac{1}{a^2} = {\left( {a + \frac{1}{a}} \right)^2} - 2\)
Final Calculation for \(a^2 + \frac{1}{a^2}\)
Using the valid value \(a + \frac{1}{a} = -7\), we can substitute this into the identity:
\(a^2 + \frac{1}{a^2} = {\left( -7 \right)^2} - 2\)
\(a^2 + \frac{1}{a^2} = 49 - 2\)
\(a^2 + \frac{1}{a^2} = 47\)
Thus, the value of \(a^2 + \frac{1}{a^2}\) is 47.
| Step | Description | Calculation |
|---|---|---|
| 1 | Given Equation | \({\left( {a + \;\frac{1}{a} + 3} \right)^2}\) = 16 |
| 2 | Take Square Root | \(a + \frac{1}{a} + 3 = \pm 4\) |
| 3 | Possible Values of \(a + \frac{1}{a}\) | 1 or -7 |
| 4 | Valid Value for Real 'a' | \(a + \frac{1}{a} = -7\) (since \({(-7)}^2 \ge 4\)) |
| 5 | Identity Used | \(a^2 + \frac{1}{a^2} = {\left( {a + \frac{1}{a}} \right)^2} - 2\) |
| 6 | Substitute Valid Value | \(a^2 + \frac{1}{a^2} = {(-7)}^2 - 2\) |
| 7 | Final Result | \(a^2 + \frac{1}{a^2} = 47\) |
Revision Table: Key Concepts for Algebraic Problems
| Concept | Description | Relevant Formula/Property |
|---|---|---|
| Square Root Property | If \(x^2 = y\), then \(x = \pm \sqrt{y}\). | \(\sqrt{k^2} = |k|\) |
| Algebraic Identity | Squaring a binomial with reciprocal terms. | \({\left( {x + \frac{1}{x}} \right)^2} = x^2 + \frac{1}{x^2} + 2\) |
| Real Number Condition for \(x + \frac{1}{x}\) | For a real number \(x \ne 0\), \(|x + \frac{1}{x}| \ge 2\). | \(x + \frac{1}{x} \ge 2\) or \(x + \frac{1}{x} \le -2\) |
Additional Information: Properties of Reciprocals and Real Numbers
The condition that 'a' is a non-zero real number is crucial in many algebraic problems. For any positive real number \(a\), \(a + \frac{1}{a} \ge 2\). This can be proven using AM-GM inequality, or by considering \(\left( \sqrt{a} - \frac{1}{\sqrt{a}} \right)^2 \ge 0\), which expands to \(a - 2 + \frac{1}{a} \ge 0\), thus \(a + \frac{1}{a} \ge 2\). Similarly, for any negative real number \(a\), let \(a = -b\) where \(b > 0\). Then \(a + \frac{1}{a} = -b + \frac{1}{-b} = -(b + \frac{1}{b})\). Since \(b > 0\), \(b + \frac{1}{b} \ge 2\). Therefore, \(a + \frac{1}{a} = -(b + \frac{1}{b}) \le -2\). Combining these, for any non-zero real number \(a\), \(|a + \frac{1}{a}| \ge 2\).
Also, for any real number \(x\), \(x^2 \ge 0\). If \(x \ne 0\), then \(x^2 > 0\) and \(\frac{1}{x^2} > 0\). The sum of two positive numbers is always positive. Thus, \(a^2 + \frac{1}{a^2} > 0\) for any non-zero real number 'a'. The value we found, 47, satisfies this condition.
Important Questions from Quadratic Equation
If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\) is:
If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:
If x 2 – 3x + 1 = 0, then the value of \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\) is:
If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) , \(x \ne 0\) , then what is the value of \((x^4+{1\over{x^2}})\over(x^2+1) \) ?
If x 2 + \(\frac{1}{x^2}\) = 18, x > 0, then find the value of x 3 + \(\frac{1}{x^3}\) .
