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Chapter 7: Problem 11

\(\sqrt{4 r+3}+1=0\)

Short Answer

Expert verified
No real solutions.

Step by step solution

01

Isolate the square root term

Subtract 1 from both sides of the equation: \[ \sqrt{4r + 3} + 1 - 1 = 0 - 1 \]This simplifies to: \[ \sqrt{4r + 3} = -1 \]
02

Understand the square root properties

Recognize that the square root of a real number cannot be negative. Since \( \sqrt{4r + 3} = -1 \) makes no sense in real numbers, there are no real solutions.

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Key Concepts

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

Isolate Square Root
To solve any equation involving a square root, the first step is to isolate the square root term. In our exercise, we start with the equation: \[ \sqrt{4r + 3} + 1 = 0 \] We need to get the square root alone on one side of the equation. To do this, we subtract 1 from both sides: \[ \sqrt{4r + 3} + 1 - 1 = 0 - 1 \] Which simplifies to: \[ \sqrt{4r + 3} = -1 \] Now, we have the square root term all by itself, making it easier to analyze and solve.
Square Root Properties
The next important thing to know is the property of square roots. The square root function, \(\sqrt{x}\), is only defined for non-negative values of \(x\). This means:
  • \sqrt{x} \geq 0 \text{ for all real numbers } \text{x} \geq 0.
In our equation, \sqrt{4r + 3} = -1,\ we have a problem. The expression on the left (\(\sqrt{4r + 3}\)) should never equal a negative number, because square roots are never negative. So, fundamentally, this equation cannot hold true for any real number \text{r}.
No Real Solutions
Understanding the square root properties helps us see why our equation has no real solutions. Since \sqrt{4r + 3}\ can never be \-1\, there is no value of \text{r}\ that would satisfy the equation: \[ \sqrt{4r + 3} = -1 \] Therefore, we conclude:
  • The original equation \sqrt{4r + 3} + 1 = 0 \text{ has no solution among real numbers.}
When solving radical equations, always remember to check whether the solution aligns with the properties of square roots. If it violates these properties, there are no real solutions.

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Most popular questions from this chapter

Simplify. Assume that all variables represent positive real numbers. \(\sqrt{121 x^{6}}\)

Find the distance between each pair of points. \((\sqrt{2}, \sqrt{6})\) and \((-2 \sqrt{2}, 4 \sqrt{6})\)

Rationalize each denominator. Assume that all variables represent positive real numbers and that no denominators are 0. $$ \frac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} $$

Rationalize each denominator. Assume that all variables represent positive real numbers and that no denominators are 0. $$ \frac{r-9}{\sqrt{r}-3} $$

Work each problem. Replace \(a\) with 3 and \(b\) with 4 to show that, in general, $$ \sqrt{a^{2}+b^{2}} \neq a+b $$
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