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Chapter 1: Problem 32

Solve by using the square root property. \((c-3)^{2}=49\)

Short Answer

Expert verified
The solutions are \( c = 10 \) and \( c = -4 \).

Step by step solution

01

Isolate the squared term

The squared term \( (c-3)^{2} \) is already isolated in the given equation \( (c-3)^{2}=49 \). So no further isolation is needed.
02

Apply the square root property

Take the square root of both sides of the equation. Remember to consider both the positive and negative roots. \[ \sqrt{(c-3)^{2}} = \pm \sqrt{49} \] This simplifies to: \[ c-3 = \pm 7 \]
03

Solve for `c`

Split the equation into two separate equations to solve for \( c \): \[ c-3 = 7 \] \[ c-3 = -7 \] Then solve both: \[(c-3) + 3 = 7 + 3 \] which gives \( c = 10 \) and \[(c-3) + 3 = -7 + 3 \] which gives \( c = -4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving quadratic equations
Let's start by understanding what quadratic equations are and how they can be solved. A quadratic equation is any equation that can be written in the form: \[ ax^2 + bx + c = 0 \] . In this format, 'a' , 'b' , and 'c' are constants, and x is the variable we need to solve for. Solving a quadratic equation means finding the values of x such that the equation holds true.
  • If a=0 , the equation reduces to a linear equation.
There are several methods to solve quadratic equations:
  • Factoring
  • Completing the square
  • Using the quadratic formula
  • Graphical method
  • Square root property
. For the given problem, we use the Square Root Property.
positive and negative roots
Quadratic equations often have two solutions because of the positive and negative roots.
When we take the square root of a number, we get both the positive and negative values. For example,
\( u^{2} = 49 \), the solutions for \( u \) are \( u = 7 \) and \( u = -7 \)
. This is why, after taking the square root of both sides of an equation, we write it as:
\( u = u_1 \text{ and } u_2 \). In our case, when we took the square root of both sides of \( (c-3)^{2}=49 \)
, we considered both \(u = 7 \) and \(u = -7 \)
isolating squared terms
Isolating the squared term is a crucial step in solving quadratic equations using the square root property. In our problem, the squared term is \((c-3)^{2} \), and it is already isolated in the given equation \((c-3)^{2}=49 \). If it weren鈥檛 isolated, we would need to perform operations such as adding, subtracting, multiplying, or dividing both sides of the equation until the squared term stands alone on one side of the equals sign.
This step simplifies the process and allows us to move directly to taking the square root of both sides.
basic algebra
Basic algebra is the foundation upon which we build our solution. After applying the square root property and getting \( c-3 = u \text{ or } u_2 \), we use basic algebra to solve for \( c \).
We isolate \( c \) by adding 3 to both sides of each of the separate equations.
  • For \( c-3 = 7 \), adding 3 gives \( c = 10 \).
  • For \( c-3 = -7 \), adding 3 yields \( c = -4 \).
These steps emphasize the importance of understanding basic algebra operations like addition and subtraction.

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Most popular questions from this chapter

Determine the set of values for \(x\) for which the radical expression would produce a real number. For example, the expression \(\sqrt{x-1}\) is a real number if \(x-1 \geq 0\) or equivalently, \(x \geq 1\). a. \(\sqrt{3 x-7}\) b. \(\sqrt[4]{3 x-7}\)

Determine the set of values for \(x\) for which the radical expression would produce a real number. For example, the expression \(\sqrt{x-1}\) is a real number if \(x-1 \geq 0\) or equivalently, \(x \geq 1\). a. \(\sqrt{x+4}\) b. \(\sqrt[3]{x+4}\)

The lengths of the sides of a triangle are given by three consecutive integers greater than \(1 .\) What are the possible values for the shortest side if the perimeter is not to exceed \(24 \mathrm{ft}\) ?

Solve the inequality. Write the solution set in interval notation. $$4 m-2 \leq-1 \quad\( or \)\quad 4 m-2 \geq 1$$

Solve the inequality, and write the solution set in interval notation. \(\left|\frac{1}{2} p-6\right| \geq 0.01\)
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