If log 10\(\rm \left[995+\sqrt{x^2-12x+60}\right]=3\) , then what is the sum of the roots of the equation ?
12
Solving the Logarithmic Equation
We are given the equation: \(\log_{10}\left[995+\sqrt{x^2-12x+60}\right]=3\)
This is a logarithmic equation. To solve for \(x\), we first need to convert the logarithmic equation into an exponential equation. The general form of a logarithmic equation is \(\log_b A = C\), which is equivalent to \(A = b^C\).
In our equation, the base \(b\) is 10, the argument \(A\) is \(995+\sqrt{x^2-12x+60}\), and the value \(C\) is 3.
So, converting the given equation into its exponential form, we get:
\(995+\sqrt{x^2-12x+60} = 10^3\)
Calculating \(10^3\):
\(10^3 = 10 \times 10 \times 10 = 1000\)
Substitute this value back into the equation:
\(995+\sqrt{x^2-12x+60} = 1000\)
Now, we need to isolate the square root term. Subtract 995 from both sides of the equation:
\(\sqrt{x^2-12x+60} = 1000 - 995\)
\(\sqrt{x^2-12x+60} = 5\)
Squaring Both Sides to Eliminate the Square Root
To get rid of the square root, we square both sides of the equation:
\(\left(\sqrt{x^2-12x+60}\right)^2 = 5^2\)
\(x^2-12x+60 = 25\)
Solving the Resulting Quadratic Equation
Now we have a quadratic equation. To solve it, we need to move all terms to one side to set the equation equal to zero. Subtract 25 from both sides:
\(x^2-12x+60 - 25 = 0\)
\(x^2-12x+35 = 0\)
This is a standard quadratic equation of the form \(ax^2+bx+c=0\), where \(a=1\), \(b=-12\), and \(c=35\). We can solve this quadratic equation by factoring or using the quadratic formula.
Let's solve it by factoring. We need to find two numbers that multiply to \(c=35\) and add up to \(b=-12\). The numbers are -7 and -5, because \((-7) \times (-5) = 35\) and \((-7) + (-5) = -12\).
So, we can factor the quadratic equation as:
\((x-7)(x-5) = 0\)
Setting each factor equal to zero gives us the roots (solutions) of the equation:
- \(x-7 = 0 \Rightarrow x = 7\)
- \(x-5 = 0 \Rightarrow x = 5\)
The roots of the quadratic equation are \(x=7\) and \(x=5\). We should verify these roots in the original equation to ensure they are valid, especially considering the square root and logarithm. For the expression under the square root, \(x^2-12x+60\), to be defined, it must be non-negative. For \(x=7\), \(7^2-12(7)+60 = 49-84+60 = 25 \ge 0\). For \(x=5\), \(5^2-12(5)+60 = 25-60+60 = 25 \ge 0\). Also, the argument of the logarithm, \(995+\sqrt{x^2-12x+60}\), must be positive. For \(x=7\), \(995+\sqrt{25} = 995+5 = 1000 > 0\). For \(x=5\), \(995+\sqrt{25} = 995+5 = 1000 > 0\). Both roots are valid.
Calculating the Sum of the Roots
The question asks for the sum of the roots of the equation. The roots we found are 7 and 5.
Sum of roots = \(7 + 5 = 12\)
Alternatively, for a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is given by the formula \(-b/a\). In our equation \(x^2-12x+35=0\), \(a=1\), \(b=-12\), and \(c=35\). The sum of the roots is:
\(\text{Sum of roots} = -\frac{b}{a} = -\frac{-12}{1} = \frac{12}{1} = 12\)
The sum of the roots is 12.
| Step | Description | Equation/Calculation |
|---|---|---|
| 1 | Convert log to exponential form | \(\log_{10}(...) = 3 \Rightarrow (...) = 10^3 = 1000\) |
| 2 | Isolate the square root | \(995 + \sqrt{...} = 1000 \Rightarrow \sqrt{...} = 5\) |
| 3 | Square both sides | \((\sqrt{x^2-12x+60})^2 = 5^2 \Rightarrow x^2-12x+60 = 25\) |
| 4 | Form a quadratic equation | \(x^2-12x+60-25 = 0 \Rightarrow x^2-12x+35=0\) |
| 5 | Find the roots (factoring) | \((x-7)(x-5) = 0 \Rightarrow x=7, x=5\) |
| 6 | Calculate the sum of roots | \(7 + 5 = 12\) |
Revision Table: Key Concepts
| Concept | Explanation | Relevance to Problem |
|---|---|---|
| Logarithmic to Exponential Conversion | \(\log_b A = C \iff A = b^C\) | Used to simplify the initial equation. |
| Solving Radical Equations | Isolate the radical term, then raise both sides to the power equal to the index of the radical (e.g., square for square root). | Used to eliminate the square root term. |
| Quadratic Equation | An equation of the form \(ax^2+bx+c=0\). | The equation simplifies to a quadratic form. |
| Sum of Roots of Quadratic | For \(ax^2+bx+c=0\), sum of roots is \(-b/a\). | Provides a direct method to find the sum of roots without explicitly finding individual roots (once the quadratic is formed). |
Additional Information: Domain of Logarithms and Square Roots
When solving equations involving logarithms and square roots, it is crucial to consider the domain of these functions:
- For \(\log_b A\), the argument \(A\) must be positive (\(A > 0\)), and the base \(b\) must be positive and not equal to 1 (\(b > 0, b \ne 1\)).
- For \(\sqrt{P}\), the expression under the square root, \(P\), must be non-negative (\(P \ge 0\)).
In this problem, the base of the logarithm is 10, which is valid. The argument of the logarithm is \(995+\sqrt{x^2-12x+60}\). For this to be defined and positive, we need \(x^2-12x+60 \ge 0\) and \(995+\sqrt{x^2-12x+60} > 0\). The quadratic \(x^2-12x+60\) has a discriminant \(\Delta = (-12)^2 - 4(1)(60) = 144 - 240 = -96 < 0\). Since the leading coefficient (1) is positive and the discriminant is negative, the quadratic \(x^2-12x+60\) is always positive for all real values of \(x\). Thus, \(x^2-12x+60 > 0\) for all \(x\), which means \(\sqrt{x^2-12x+60}\) is always a real, positive value. Consequently, \(995 + \sqrt{x^2-12x+60}\) is always positive, satisfying the logarithm's domain requirement. Both roots \(x=7\) and \(x=5\) are therefore valid solutions to the original equation.
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