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Find an exact solution to each equation. (Leave your answers in radical form.) a. \(x^{2}=47\) b. \((x-4)^{2}=28\) c. \((x+2)^{2}-3=11\) d. \(2(x-1)^{2}+4=18\)

Short Answer

Expert verified
a. \(\pm\sqrt{47}\); b. \(4 \pm 2\sqrt{7}\); c. \(-2 \pm \sqrt{14}\); d. \(1 \pm \sqrt{7}\).

Step by step solution

01

Solve equation a

Given equation: \(x^2 = 47\). We take the square root of both sides to solve for \(x\). This gives us: \[x = \pm\sqrt{47}.\] Since 47 is not a perfect square, the answer remains in radical form.
02

Solve equation b

Given equation: \((x - 4)^2 = 28\). We take the square root of both sides to solve for \(x - 4\). This gives us: \[x - 4 = \pm\sqrt{28}.\] Simplify \(\sqrt{28}\) to \(\sqrt{4 \times 7} = 2\sqrt{7}\). So, \(x - 4 = \pm 2\sqrt{7}\). Finally, solve for \(x\): \[x = 4 \pm 2\sqrt{7}.\]
03

Solve equation c

Given equation: \((x + 2)^2 - 3 = 11\). First, add 3 to both sides to isolate the squared term: \((x + 2)^2 = 14\). Next, take the square root of both sides: \[x + 2 = \pm\sqrt{14}.\] Finally, solve for \(x\): \[x = -2 \pm \sqrt{14}.\]
04

Solve equation d

Given equation: \(2(x - 1)^2 + 4 = 18\). First, subtract 4 from both sides to isolate the term with \(x\): \(2(x - 1)^2 = 14\).Divide by 2: \((x - 1)^2 = 7\).Now, take the square root of both sides: \[x - 1 = \pm\sqrt{7}.\] Finally, solve for \(x\): \[x = 1 \pm \sqrt{7}.\]

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Key Concepts

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

Solving Quadratic Equations
Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. Solving these equations involves finding the values of \(x\) that satisfy the equation. There are several methods to solve quadratic equations including:
  • Factoring: This method involves writing the quadratic equation as a product of two linear equations.
  • Completing the square: By manipulating the equation, we form a perfect square trinomial on one side, simplifying the solution process.
  • Quadratic formula: This is a reliable method, using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Taking square roots: Applicable when the equation can be rearranged into the form \((expression)^2 = constant\).
In our exercises, we primarily use completing the square and taking square roots to find the exact solutions. Understanding these approaches enables tackling various kinds of quadratic equations.
Radical Expressions
Radical expressions involve roots, notably square roots, which occur quite often in quadratic solutions. A radical expression includes the radical symbol \(\sqrt{}\) and a radicand, which is the value or expression inside the radical sign. Let's consider an example: if dealing with \(\sqrt{28}\), from one of our exercises, breaking it down into its prime factors (i.e., \(28 = 4 \times 7\)) helps in simplifying to \(2\sqrt{7}\). Simplification is crucial for clarity and simplifying further operations. Understanding radicals is fundamental when solving quadratic equations, as they often appear during the process when extracting square roots from both sides of an equation.
Square Roots
A square root implies a value that, when multiplied by itself, results in the original number. For example, the square root of \(25\) is \(5\) (since \(5 \times 5 = 25\)). When dealing with quadratic equations, taking square roots is a common step, especially when equations are converted into the form \(x^2 = n\). In this form, solving for \(x\) requires computing \(x = \pm \sqrt{n}\), showing both the positive and negative roots are solutions. For instance, in the equation \(x^2 = 47\), solving involves calculating \(x = \pm\sqrt{47}\). Note that square roots of non-perfect squares, like \(47\), remain in this radical form as simplifying further isn't feasible. The ability to handle square roots, both perfect and imperfect, is essential for efficiently working through quadratic problems.
Factoring
Factoring in algebra refers to expressing a mathematical expression as a product of simpler factors or roots. In the context of quadratic equations, factoring breaks down the equation \(ax^2 + bx + c = 0\) into a product of binomials. However, in our provided problems, direct factoring wasn't heavily utilized because the equations didn't neatly fit typical factorable patterns. Despite this, understanding factorization is vital. Suppose an equation like \(x^2 + 5x + 6 = 0\). We factor it into \((x + 2)(x + 3) = 0\), leading to solutions \(x = -2\) or \(x = -3\). Recognizing when to employ factoring is important, especially when equations feature easily identifiable patterns or when seeking to verify different solving strategies.

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