Select whether the equation has a solution or not.
[tex]\sqrt{4 x+5}-\sqrt{x-1}=\sqrt{x+4}[/tex]
A. no roots
B. roots
Answer (2)
Find the domain of the equation: x ≥ 1 .
Square both sides of the equation and simplify to get 2 x = 4 x 2 + x − 5 .
Square both sides again to obtain 4 x 2 = 4 x 2 + x − 5 , which simplifies to x = 5 .
Check if the solution x = 5 satisfies the domain and the original equation. Since it does, the equation has a solution: roo t s .
Explanation
Problem Analysis We are given the equation 4 x + 5 − x − 1 = x + 4 and asked to determine if it has a solution.
Finding the Domain First, let's find the domain of the equation. Since we have square roots, we need to ensure that the expressions inside the square roots are non-negative:
4 x + 5 ≥ 0 ⇒ x ≥ − 4 5
x − 1 ≥ 0 ⇒ x ≥ 1
x + 4 ≥ 0 ⇒ x ≥ − 4
Combining these inequalities, we find that the domain of the equation is x ≥ 1 .
Squaring Both Sides (First Time) Now, let's solve the equation. Square both sides of the equation:
( 4 x + 5 − x − 1 ) 2 = ( x + 4 ) 2
( 4 x + 5 ) − 2 ( 4 x + 5 ) ( x − 1 ) + ( x − 1 ) = x + 4
5 x + 4 − 2 4 x 2 − 4 x + 5 x − 5 = x + 4
5 x + 4 − 2 4 x 2 + x − 5 = x + 4
Isolating the Square Root Isolate the square root term:
4 x = 2 4 x 2 + x − 5
2 x = 4 x 2 + x − 5
Squaring Both Sides (Second Time) Square both sides again:
( 2 x ) 2 = ( 4 x 2 + x − 5 ) 2
4 x 2 = 4 x 2 + x − 5
0 = x − 5
x = 5
Checking the Solution Now, we need to check if the solution x = 5 is within the domain x ≥ 1 . Since 5 ≥ 1 , the solution is within the domain.
Next, we need to check if x = 5 satisfies the original equation:
4 ( 5 ) + 5 − 5 − 1 = 5 + 4
25 − 4 = 9
5 − 2 = 3
3 = 3
The solution x = 5 satisfies the original equation.
Conclusion Therefore, the equation has a solution, and that solution is x = 5 .
Examples
When designing a bridge, engineers use equations involving square roots to calculate the tension and compression forces in the supporting cables and structures. Solving these equations ensures the bridge's stability and safety under various loads. The ability to manipulate and solve such equations is crucial in ensuring structural integrity and preventing potential failures.
The equation s q r t 4 x + 5 − s q r t x − 1 = s q r t x + 4 has a domain of x ≥ 1 . After checking potential solutions, we find that the only candidate does not satisfy the original equation. Therefore, the equation has no roots.
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