Problem 81
Question
For the following problems, simplify each of the radical expressions. $$ \sqrt{a^{2}-10 a+25} $$
Step-by-Step Solution
Verified Answer
Based on the step-by-step solution, provide a short answer:
To simplify the radical expression \(\sqrt{a^{2} - 10 a + 25}\), we factor the quadratic polynomial inside the square root and simplify by canceling out the square root and square operations. The simplified form of the given expression is \(a - 5\).
1Step 1: Factor the quadratic polynomial inside the square root
We have the given expression as
$$
\sqrt{a^{2}-10 a+25}
$$
Let's try factoring the polynomial inside the square root:
$$
a^2 - 10a + 25 = (a - 5)(a - 5)
$$
So the expression becomes:
$$
\sqrt{(a - 5)(a - 5)}
$$
2Step 2: Simplify the square root
Now that we have factored the polynomial inside the square root, we can simplify the square root:
$$
\sqrt{(a - 5)(a - 5)} = \sqrt{(a - 5)^2}
$$
As the square root and square operations are inverse of each other, we can cancel them out:
$$
\sqrt{(a - 5)^2} = a - 5
$$
3Step 3: Present the simplified expression
The simplified expression is:
$$
a - 5
$$
Therefore, the simplified form of the given radical expression is \(a - 5\).
Key Concepts
Factor Quadratic PolynomialSquare Root SimplificationInverse Operations
Factor Quadratic Polynomial
When faced with simplifying radical expressions containing a quadratic polynomial, the first crucial step is factoring. Factoring transforms the quadratic polynomial into a product of simpler expressions, often binomials or monomials, making it easier to work with.
To factor a quadratic polynomial, like \(a^{2} - 10a + 25\), we look for two numbers that multiply to give the constant term (in this case 25) and add to give the coefficient of the linear term (here, -10). When factoring perfect square trinomials, these numbers are the same, leading to a binomial squared such as \((a - 5)^{2}\).
To factor a quadratic polynomial, like \(a^{2} - 10a + 25\), we look for two numbers that multiply to give the constant term (in this case 25) and add to give the coefficient of the linear term (here, -10). When factoring perfect square trinomials, these numbers are the same, leading to a binomial squared such as \((a - 5)^{2}\).
Why Factor a Quadratic?
Factoring can simplify complex equations, reveal hidden relationships, and even shed light on the graph of the function, such as its roots or its vertex. When a quadratic polynomial is inside a square root, factoring is essential as it can often lead to square root simplification by revealing perfect squares.Square Root Simplification
Square root simplification is a common technique in algebra that involves finding an exact value or a simpler form of a square root expression.
When we simplify \(\sqrt{(a - 5)^2}\), we are looking for a number that, when multiplied by itself, gives the original value under the radical. Since squaring a number and taking the square root are inverse operations, we can simplify \((a - 5)^2\) by simply removing these operations and returning the value inside the parentheses, which is \(a - 5\).
When we simplify \(\sqrt{(a - 5)^2}\), we are looking for a number that, when multiplied by itself, gives the original value under the radical. Since squaring a number and taking the square root are inverse operations, we can simplify \((a - 5)^2\) by simply removing these operations and returning the value inside the parentheses, which is \(a - 5\).
Dealing with Negative Numbers
Remember that when simplifying square roots, if the value inside the square is negative, the result will involve imaginary numbers because the square root of a negative number is not a real number. However, in the context of real numbers, we simplify under the assumption that the variable within the square root leads to a non-negative result.Inverse Operations
Inverse operations are pairs of mathematical operations that undo each other. The most common pairs include addition with subtraction and multiplication with division.
For simplification purposes, recognizing inverse operations helps in solving equations and simplifying expressions, such as radical expressions. In the example \(\sqrt{(a - 5)^2}\), squaring a number and taking the square root are inverse operations. Applying them sequentially to a number returns us to the original number. This is why when we encounter \(\sqrt{(a - 5)^2}\), the two operations cancel each other and we are left with just the value of \(a - 5\).
For simplification purposes, recognizing inverse operations helps in solving equations and simplifying expressions, such as radical expressions. In the example \(\sqrt{(a - 5)^2}\), squaring a number and taking the square root are inverse operations. Applying them sequentially to a number returns us to the original number. This is why when we encounter \(\sqrt{(a - 5)^2}\), the two operations cancel each other and we are left with just the value of \(a - 5\).
The Power of Inverse Operations
Understanding inverse operations is powerful for checking work, solving complicated equations, and simplifying expressions. By applying an inverse operation, we can isolate variables, undo complex mathematical processes, and in many cases, directly determine the solution to a problem.Other exercises in this chapter
Problem 81
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Perform the indicated operations: $$ \frac{15 x^{2}-20 x}{6 x^{2}+x-12} \cdot \frac{8 x+12}{x^{2}-2 x-15} \div \frac{5 x^{2}+15 x}{x^{2}-25} $$
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