Problem 50
Question
Solve using the square root property. Simplify all radicals. $$ (x-7)^{2}=16 $$
Step-by-Step Solution
Verified Answer
x = 11 or x = 3
1Step 1 - Isolate the squared term
Notice that the equation \[ (x-7)^2 = 16 \] already has the squared term isolated.
2Step 2 - Apply the square root property
Apply the square root property to both sides of the equation, which states that if \[ a^2 = b \] then \[ a = \pm \sqrt{b} \]. Thus: \[ \sqrt{(x-7)^2} = \pm \sqrt{16} \] This simplifies to: \[ x-7 = \pm 4 \]
3Step 3 - Solve for x
Now solve the two resulting equations: \[ x-7 = 4 \] and \[ x-7 = -4 \]. For \[ x-7 = 4: \] \[ x = 4 + 7 \] \[ x = 11 \] For \[ x-7 = -4: \] \[ x = -4 + 7 \] \[ x = 3 \]
Key Concepts
RadicalsIsolating VariablesSolving Quadratic Equations
Radicals
Radicals, also known as square roots, are an essential concept when working with equations, especially quadratic equations. Radicals are represented by the square root symbol \(\backslash \sqrt{}\) and are used to find a number which, when multiplied by itself, gives the original value inside the radical.
For example, \( \sqrt{16} = 4 \) because \( 4 \cdot 4 = 16 \).
When dealing with radical expressions in equations, always remember to simplify them where possible. This helps in getting the exact value and makes further calculations easier. In the exercise given, simplifying radicals was necessary to solve the equation \( (x - 7)^2 = 16 \), leading to \( \sqrt{(x-7)^2} = \pm \sqrt{16} \). This simplifies to \( x - 7 = \pm 4 \).
Notice how simplifying radicals was essential to solve for \( x \) in this context. Simplification steps can often make complex problems much more manageable.
For example, \( \sqrt{16} = 4 \) because \( 4 \cdot 4 = 16 \).
When dealing with radical expressions in equations, always remember to simplify them where possible. This helps in getting the exact value and makes further calculations easier. In the exercise given, simplifying radicals was necessary to solve the equation \( (x - 7)^2 = 16 \), leading to \( \sqrt{(x-7)^2} = \pm \sqrt{16} \). This simplifies to \( x - 7 = \pm 4 \).
Notice how simplifying radicals was essential to solve for \( x \) in this context. Simplification steps can often make complex problems much more manageable.
Isolating Variables
Isolating variables is a fundamental skill in algebra used to solve equations. It involves manipulating the equation to get the variable you are solving for alone on one side of the equation.
In our exercise, the variable \( x \) needed to be isolated to find its value. The initial equation was \( (x - 7)^2 = 16 \). By applying the square root property, it became \( x - 7 = \pm 4 \). Here, isolating \( x \) required simple algebraic steps:
Isolating variables often includes adding, subtracting, multiplying, or dividing both sides of the equation by the same number. These operations help in eliminating other terms around the variable, making it easier to solve.
In our exercise, the variable \( x \) needed to be isolated to find its value. The initial equation was \( (x - 7)^2 = 16 \). By applying the square root property, it became \( x - 7 = \pm 4 \). Here, isolating \( x \) required simple algebraic steps:
- For \( x - 7 = 4 \), adding 7 to both sides gave \( x = 11 \)
- For \( x - 7 = -4 \), adding 7 to both sides gave \( x = 3 \)
Isolating variables often includes adding, subtracting, multiplying, or dividing both sides of the equation by the same number. These operations help in eliminating other terms around the variable, making it easier to solve.
Solving Quadratic Equations
Solving quadratic equations is a critical part of algebra. A typical quadratic equation is in the form \( ax^2 + bx + c = 0 \). There are multiple methods to solve such equations, including factoring, completing the square, and using the quadratic formula. In some cases, like our exercise \( (x - 7)^2 = 16 \), we can use the square root property.
The square root property is particularly useful when the quadratic equation can be expressed in the form \( (x - p)^2 = q \). It states that if \((a)^2 = b\), then \(a = \pm \sqrt{b} \).
Applying this to \( (x - 7)^2 = 16 \), we get:
\(\backslash \sqrt{(x-7)^2} = \pm \sqrt{16} \)
This simplifies to \( x - 7 = \pm 4 \), providing two linear equations:
The square root property is particularly useful when the quadratic equation can be expressed in the form \( (x - p)^2 = q \). It states that if \((a)^2 = b\), then \(a = \pm \sqrt{b} \).
Applying this to \( (x - 7)^2 = 16 \), we get:
\(\backslash \sqrt{(x-7)^2} = \pm \sqrt{16} \)
This simplifies to \( x - 7 = \pm 4 \), providing two linear equations:
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Problem 49
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