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.

Revision Table: Key Concepts

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.