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Chapter 7: Problem 12

Solve each equation.. \((d+7)^{2}=0\)

Short Answer

Expert verified
The short answer to the equation \((d+7)^2 = 0\) is \(d = -7\).

Step by step solution

01

Write down the given equation

Begin by writing down the equation provided: \((d+7)^2 = 0\).
02

Expand the square

Now, expand the square by multiplying \((d+7)\) by itself: \((d+7)(d+7) = 0\).
03

Use the distributive property

Using the distributive property, multiply each term inside the first parentheses by each term inside the second parentheses: \(d \cdot d + d \cdot 7 + 7 \cdot d + 7 \cdot 7 = 0\) This simplifies to: \(d^2 + 7d + 7d + 49 = 0\)
04

Combine like terms

Next, combine the like terms (7d and 7d) in the equation: \(d^2 + 14d + 49 = 0\)
05

Solve for d

Since the equation is a quadratic equation, we can try to factor it to find the value of d that makes the equation true. However, in this case, we can notice that the equation is already a perfect square: \((d + 7)^2 = 0\) Therefore, we can simply set the term inside the parentheses equal to 0: \(d + 7 = 0\) Now, subtract 7 from both sides of the equation to solve for d: \(d = -7\)
06

Check the solution

To ensure that our solution is correct, we can plug the value of d back into the original equation and check if it holds true: \((d + 7)^2 = (-7 + 7)^2 = (0)^2 = 0\) Since the equation holds true, the solution is correct: \(d = -7\).

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Key Concepts

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

Distributive Property
The distributive property is a fundamental principle in algebra that is essential for simplifying expressions and equations. It allows us to expand expressions like
  • \((a+b)(c+d)\) into \(ac+ad+bc+bd\)
  • ensures that each term in one parentheses is multiplied by each term in the other.
This property is applied in solving our given equation, \((d+7)^2 = 0\), when expanding and simplifying the expression to get
  • \(d^2 + 7d + 7d + 49 = 0\).
These steps ensure that the square of the binomial is correctly expanded by distributing each term of
  • \((d+7)\) across \((d+7)\).
With this property, you can break down complex algebraic expressions into more manageable parts, making it easier to solve equations.
Factoring Quadratic Equations
Factoring quadratic equations is a technique used to solve equations of the form
  • \(ax^2 + bx + c = 0\).
It's a handy method when the quadratic can be written as
  • \((x+p)(x+q) = 0\),
allowing us to find solutions by setting each factor equal to zero:
  • \(x+p=0\) or \(x+q=0\).
For our specific problem, the equation \((d+7)^2=0\) is already a perfect square. This means instead of factoring further, we recognize
  • \((d+7)(d+7)=0\)
  • implies \(d+7=0\).
Often with factored quadratics, you'd solve each factor separately. Here, however, the symmetry means both solutions are the same:
  • \(d = -7\).
Combining Like Terms
Combining like terms is a simplification process in algebra where you consolidate terms with identical variables and powers. This method streamlines expressions, reducing them to simpler forms. For example, terms like
  • \(7d + 7d\)
can be combined by adding their coefficients, resulting in
  • \(14d\).
In solving our quadratic equation, combining like terms transformed
  • \(d^2 + 7d + 7d + 49 = 0\)
  • into \(d^2 + 14d + 49 = 0\).
This reduction clarifies the structure of the equation, making it easier to notice shortcuts such as factoring or recognizing perfect squares. Efficiently combining like terms is crucial for dealing with longer and more complex algebraic expressions.

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Most popular questions from this chapter

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$(3 x-1)\left(x^{2}-16 x+64\right)=0$$

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$10 n^{2}(n-8)+n(n-8)-2(n-8)=0$$

Factor completely. $$d^{3}+1$$

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$(2 r-5)\left(r^{2}-6 r+9\right)=0$$

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$8 y(y+4)(2 y-1)=0$
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