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Chapter 1: Problem 36

Is this reasoning for finding the solutions of the equation \(\sqrt{2 x^{2}-1}=x\) correct? \((1) \sqrt{2 x^{2}-1}=x\) is given; (2) \(2 x^{2}-1=x^{2},\) obtained by squaring both sides of \((1)\) (3) \(x^{2}-1=0,\) obtained by subtracting \(x^{2}\) from both sides of \((2) ;(4)(x-1)(x+1)=0\) , obtained by factoring the left-hand side of \(x^{2}-1 ;\) (5) \(x=1\) or \(x=-1\) which follows because \(a b=0\) implies that \(a=0\) or \(b=0\)

Short Answer

Expert verified
The reasoning is partially correct. Only the solution \(x = 1\) is valid.

Step by step solution

01

Identify the Given Equation

The given equation is \(\sqrt{2x^2 - 1} = x\).
02

Square Both Sides

Square both sides of the equation to remove the square root. This gives: \[ (\sqrt{2x^2 - 1})^2 = x^2 \] which simplifies to \[ 2x^2 - 1 = x^2 \].
03

Simplify the Equation

Subtract \(x^2\) from both sides to combine like terms: \[ 2x^2 - 1 - x^2 = 0 \] which simplifies to \[ x^2 - 1 = 0 \].
04

Factor the Equation

Factor the left-hand side of the equation \(x^2 - 1 = 0\). This can be written as: \[ (x - 1)(x + 1) = 0 \].
05

Find the Solutions

Set each factor equal to zero and solve for \(x\): \[ x - 1 = 0 \implies x = 1 \] and \[ x + 1 = 0 \implies x = -1 \].
06

Verify the Solutions

Check both solutions in the original equation \(\sqrt{2x^2 - 1} = x\). For \(x = 1\), \(\sqrt{2(1)^2 - 1} = \sqrt{2 - 1} = 1\), which is true. For \(x = -1\), \(\sqrt{2(-1)^2 - 1} = \sqrt{2 - 1} = 1\), but \(-1 eq 1\). Hence, \(x = -1\) is not a valid solution.

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Key Concepts

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

Square Roots
Square roots are incredibly important when solving equations. In the problem at hand, dealing with the equation \(\text{\sqrt\{2x^2 - 1\}} = x\), the square root function helps us to simplify complex expressions. To solve such equations, you first need to eliminate the square root by squaring both sides. This means you will convert \((\sqrt\{2x^2 - 1\})^2\) into a polynomial \(2x^2 - 1\). Remember this simple trick: squaring both sides helps you get rid of the square root and makes the equation easier to solve.
Factoring Polynomials
Factoring polynomials is all about finding expressions that multiply together to give the original polynomial. In this exercise, after squaring and rearranging terms, you get the polynomial \(x^2 - 1\). This can be factored into \((x - 1)(x + 1) = 0\). Factoring makes it easy to find the solutions to the equation. It splits the polynomial into simpler parts, allowing you to set each factor to zero and solve for \(x\) separately.
Verifying Solutions
Once you obtain potential solutions, you must verify them in the original equation to ensure they truly satisfy it. After finding \(x = 1\) and \(-1\), substituting these back into \(\sqrt{2x^2 - 1} = x\) ensures that only \(x = 1\) is valid. This step is crucial in solving equations, as squaring both sides can sometimes introduce extraneous solutions that don't satisfy the original equation.
Discrete Mathematics
Discrete mathematics plays a key role in understanding and solving equations, particularly through logical reasoning and proof strategies. When you solve for \(x\) and check each potential solution, you're performing a step of logical validation. Discrete math often requires you to ensure each solution meets all criteria originally set, as we did by verifying \(-1\) didn't fulfill the initial equation. This guarantees the accuracy and correctness of the solution you present.

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Most popular questions from this chapter

Prove that there is no positive integer \(n\) such that \(n^{2}+\) \(n^{3}=100 .\)

Prove that there are 100 consecutive positive integers that are not perfect squares. Is your proof constructive or non-constructive?

Suppose that the domain of the propositional function \(P(x)\) consists of the integers \(1,2,3,4,\) and \(5 .\) Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. $$ \begin{array}{ll}{\text { a) } \quad \exists x P(x)} & {\text { b) } \forall x P(x)} \\ {\text { c) } \quad \neg \exists x P(x)} & {\text { d) } \neg \forall x P(x)}\end{array} $$ e) \(\quad \forall x((x \neq 3) \rightarrow P(x)) \vee \exists x \neg P(x)\)

Construct a truth table for each of these compound propositions. a) \(p \rightarrow(\neg q \vee r)\) b) \(\neg p \rightarrow(q \rightarrow r)\) c) \((p \rightarrow q) \vee(\neg p \rightarrow r)\) d) \((p \rightarrow q) \wedge(\neg p \rightarrow r)\) e) \((p \rightarrow q) \vee(\neg q \rightarrow r)\) f) \((\neg p \leftrightarrow \neg q) \leftrightarrow(q \leftrightarrow r)\)

Determine whether these are valid arguments. a) If \(x\) is a positive real number, then \(x^{2}\) is a positive real number. Therefore, if \(a^{2}\) is positive, where \(a\) is a real number, then \(a\) is a positive real number. b) If \(x^{2} \neq 0,\) where \(x\) is a real number, then \(x \neq 0 .\) Let \(a\) be a real number with \(a^{2} \neq 0 ;\) then \(a \neq 0\)
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