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Chapter 9: Problem 15

Solve the quadratic equations in Exercises 11-22 by taking square roots. $$ (x-3)^{2}-6=10 $$

Short Answer

Expert verified
Answer: The solutions for the quadratic equation (x-3)^2 - 6 = 10 are x = 7 and x = -1.

Step by step solution

01

Rearrange the equation

Add 6 to both sides of the equation: \((x-3)^2 - 6 + 6 = 10 + 6\) This simplifies to: \((x-3)^2 = 16\)
02

Take the square root of both sides

Now we will take the square root of both sides of the equation: \(\sqrt{(x-3)^2} = \sqrt{16}\)
03

Simplify the square root

The square root of \((x-3)^2\) is \(x-3\), and the square root of \(16\) is \(4\). We also have to consider the negative square root of \(16\), which is \(-4\). We write it as: \(x - 3 = \pm4\)
04

Solve for x

Now we will solve for x by adding 3 to both sides of the equation: \(x - 3 + 3 = \pm4 + 3\) This simplifies to: \(x = \pm4 + 3\) We find two solutions for x: 1. \(x = 4 + 3 = 7\) 2. \(x = -4 + 3 = -1\) So, the solutions for the quadratic equation \((x-3)^2 - 6 = 10\) are \(x = 7\) and \(x = -1\).

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Key Concepts

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

Solving by Square Roots
Solving a quadratic equation by square roots is a straightforward technique for equations that have a squared term set equal to a number. This approach is particularly useful when the quadratic equation is in the form
  • \((x-a)^2 = b\)
To solve such an equation, your first step is to isolate the squared term by performing algebraic operations. In the example above,
  • \((x-3)^2 = 16\)
was achieved after adding 6 to both sides. Next, take the square root of both sides. Remember, every positive real number has two square roots: a positive and a negative. Therefore, the solution step is
  • \(x - 3 = \pm 4\)
Finally, solve for the variable \(x\) by isolating it using basic arithmetic operations, resulting in two possible solutions.
Algebraic Manipulation
Algebraic manipulation is a key element in solving quadratic equations. It involves rearranging and simplifying equations to isolate desired terms. For the equation
  • \((x-3)^2 - 6 = 10\)
you start by moving the constant away from the squared term. Adding 6 to both sides simplifies the equation to
  • \((x-3)^2 = 16\)
Algebraic manipulation doesn't change the equality; it merely rearranges it, so you can apply other mathematical principles, like taking a square root, effectively.
Remember, each operation should aim to isolate the variable or simplify the equation.
This requires understanding properties of operations, such as the additive and multiplicative identities, as used in the steps provided.
Quadratic Solutions
Quadratic equations can be solved using various methods, with solutions potentially being real numbers, complex numbers, or repeated roots. In our example, the solutions are real numbers.
  • The equation \((x-3)^2 - 6 = 10\) leads us to the standard quadratic form first.
By manipulating the equation through the square root method, the solutions turn out to be \(x = 7\) and \(x = -1\). These solutions indicate the values of \(x\) for which the original equation holds true.
Quadratic solutions give insight into the function's behavior. Remember, solving quadratic equations by square roots is one of many methods, but it is particularly efficient for equations that can be easily rearranged to reveal a perfect square on one side.

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Most popular questions from this chapter

Solve by (a) Completing the square (b) Using the quadratic formula $$ 2 x^{2}-32 x+7=0 $$

Use what you know about the quadratic formula to find a quadratic equation having $$ x=\frac{-2 \pm \sqrt{8}}{2} $$ as solutions. Your equation should be in standard form with integer (whole number) coefficients.

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Find a quadratic function with the given zeros and write it in standard form. 3 and 4

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