Edulastic Formative and Summative Assessment app.edulastic.com/student/assessment/5f882237e806f400084dag3f/class/5f4obeb5sdd6ded177ade83/uta/5f8828493130910008e4a9b6/qid/13 Question 14/17 NEXT BOOKMARK Solve the following equation for (List solutions from least to greatest: If an answer box is not needed, write “none” in that box) -1 = √(V@+n) + 4 Valid Solutions:
The Correct Answer and Explanation is:
We are given the equation:−1=V@+n+4-1 = \sqrt{V@ + n} + 4−1=V@+n+4
It looks like the original equation contains a typographical error: “√(V@ + n)” is not valid mathematical notation. Let’s assume the correct equation is something more standard, such as:−1=x+n+4-1 = \sqrt{x + n} + 4−1=x+n+4
Our goal is to solve the equation for xxx and list all valid solutions from least to greatest. If we assume that nnn is a known constant (say, n=5n = 5n=5), then the equation becomes:−1=x+5+4-1 = \sqrt{x + 5} + 4−1=x+5+4
Step-by-step Solution:
Step 1: Isolate the square root expression
Subtract 4 from both sides:−1−4=x+5⇒−5=x+5-1 – 4 = \sqrt{x + 5} \Rightarrow -5 = \sqrt{x + 5}−1−4=x+5⇒−5=x+5
Step 2: Analyze the square root expression
Here, we have:x+5=−5\sqrt{x + 5} = -5x+5=−5
This equation has no solution, because the square root of a real number is always non-negative (i.e., a≥0\sqrt{a} \geq 0a≥0 for any real number a≥0a \geq 0a≥0). It can never equal a negative number like -5.
Conclusion:
There is no real solution to this equation. The square root function cannot produce a negative output, so the equation is not valid for any real value of xxx.
Final Answer:
Valid Solutions:
none
Explanation
To solve a radical equation, our first step is to isolate the radical (the square root in this case). The given equation is:−1=x+5+4-1 = \sqrt{x + 5} + 4−1=x+5+4
We begin by isolating the square root on one side. This is achieved by subtracting 4 from both sides:−1−4=x+5⇒−5=x+5-1 – 4 = \sqrt{x + 5} \Rightarrow -5 = \sqrt{x + 5}−1−4=x+5⇒−5=x+5
Now we examine the resulting expression. The square root function x+5\sqrt{x + 5}x+5 represents the principal (non-negative) square root. This means it can only yield values greater than or equal to zero. In mathematical terms:x+5≥0\sqrt{x + 5} \geq 0x+5≥0
However, in our equation, the square root is set equal to -5. This is not possible, because there is no real number whose square root is negative. Therefore, the equation:−5=x+5-5 = \sqrt{x + 5}−5=x+5
is not valid for any real value of xxx. Attempting to solve such an equation leads to a contradiction of the basic properties of square roots.
Hence, the conclusion is that the equation has no solution in the set of real numbers. In formal math notation, we might write:No Solutionornone\boxed{\text{No Solution}} \quad \text{or} \quad \boxed{\text{none}}No Solutionornone
This highlights an important concept: Always check whether the operations in an equation are defined for real numbers. In this case, the equation fails that test due to the square root equaling a negative number
