If log 10$\rm \left[995+\sqrt{x^2-12x+60}\right]=3$ , then what is the sum of the roots of the equation ?

Question

Prepp Team · Accepted Answer

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.

Prepp Team · Answer

11

Prepp Team · Answer

10

Prepp Team · Answer

9

Solving the Logarithmic Equation

Squaring Both Sides to Eliminate the Square Root

Solving the Resulting Quadratic Equation

Calculating the Sum of the Roots

Revision Table: Key Concepts

Additional Information: Domain of Logarithms and Square Roots

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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	\(7 + 5 = 12\)

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).