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Chapter 11: Problem 52

Solve using the square root property. Simplify all radicals. $$ (x+3)^{2}=11 $$

Short Answer

Expert verified
x = -3 \pm \sqrt{11}

Step by step solution

01

- Isolate the Squared Term

The squared term \( (x+3)^{2} \) is already isolated on the left-hand side of the equation \( (x+3)^{2}=11 \).
02

- Apply the Square Root Property

Take the square root of both sides of the equation to eliminate the square. Remember to include both the positive and negative square roots: \[ \sqrt{(x+3)^{2}} = \pm \sqrt{11} \Rightarrow x+3 = \pm \sqrt{11} \]
03

- Solve for x

To isolate x, subtract 3 from both terms on the right-hand side: \[ x = -3 \pm \sqrt{11} \]
04

- Simplify the Radicals

The \sqrt{11} is already in its simplest form, so your final simplified solutions for x are: \[ x = -3 + \sqrt{11} \] \[ x = -3 - \sqrt{11} \]

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Key Concepts

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

square root property
The square root property is an essential tool for solving quadratic equations. It states that if we have an equation in the form \(x^2 = k\), we can solve it by taking the square root of both sides. This gives us two potential solutions, since the square root of a number can be positive or negative. For example, \(x^2 = 11\) leads to \(x = \pm\sqrt{11}\). In our original exercise, we apply this property to \( (x+3)^2 = 11 \) to find \( x+3 = \pm\sqrt{11} \). Always remember to include both positive and negative roots when applying the square root property.
isolating the variable
Isolating the variable is a crucial step in solving equations. This means getting the variable by itself on one side of the equation. For our exercise, we start with \( (x+3)^2 = 11 \). Luckily, the squared term \( (x+3)^2 \) is already isolated. After taking the square root of both sides, \( x+3 = \pm\sqrt{11} \), we further isolate x by subtracting 3 from both sides. This leaves us with \( x = -3 \pm\sqrt{11} \). Properly isolating the variable ensures we correctly solve for the unknown.
simplifying radicals
Simplifying radicals means expressing the radical in its simplest form. In our problem, \(\sqrt{11}\) is already simplified, as 11 is a prime number and has no square factors other than 1. Take, for example, \( \sqrt{12} \): we can break it down into \(\sqrt{4 \times 3} \), which simplifies further to \ 2\sqrt{3}\. Simplifying radicals makes them easier to work with in later steps and ensures your final answer is in its most reduced form.
positive and negative roots
When solving equations using the square root property, we must always consider both positive and negative roots. For instance, in the problem \( (x+3)^2 = 11 \), taking the square root of both sides gives us \( x+3 = \pm\sqrt{11} \). This results in two solutions: \ x+3 = \sqrt{11} \ and \ x+3 = -\sqrt{11} \. Solving each, we find \ x = -3 + \sqrt{11} \ and \ x = -3 - \sqrt{11}\. Including both roots is essential for capturing all possible solutions, as neglecting one might lead to an incomplete answer.

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Most popular questions from this chapter

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (x-1)(9 x-3)=-2 $$

Find the discriminant. Use it to determine whether the solutions for each equation are A. two rational numbers B. one rational number C. two irrational numbers D. two nonreal complex numbers. Tell whether the equation can be solved using the zero-factor property, or if the quadratic formula should be used instead. Do not actually solve. $$ 4 x^{2}=4 x+3 $$

Solve each problem using a quadratic equation. Use the formula \(A=P(1+r)^{2}\) to find the interest rate \(r\) at which a principal \(P\) of \(\$ 10,000\) will increase to \(\$ 10,920.25\) in 2 yr.

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A model rocket is projected vertically upward from the ground. Its distance \(s\) in feet above the ground after t seconds is given by the quadratic function $$ s(t)=-16 t^{2}+256 t $$ Work Exercise in order, to see how quadratic equations and inequalities are related. At what times will the rocket be at ground level? (Hint: Let \(s(t)=0\) and solve the quadratic equation.)
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