How many real solutions does the equation $\sqrt{3 x-1}=2 x-5$ have?
A. 0
B. 1
C. 2
D. cannot be determined from the graph
Answer (2)
The equation 3 x − 1 = 2 x − 5 has only one real solution. After determining the domain and checking for extraneous solutions, we find that only one value satisfies all conditions. Thus, the answer is B. 1 .
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Square both sides of the equation to get rid of the square root: 3 x − 1 = ( 2 x − 5 ) 2 .
Simplify and rearrange the equation into a quadratic form: 4 x 2 − 23 x + 26 = 0 .
Solve the quadratic equation using the quadratic formula to find two potential solutions: x 1 ≈ 1.546 and x 2 ≈ 4.204 .
Check for extraneous solutions by verifying that x ≥ 2 5 . Only x 2 satisfies this condition, so the final answer is 1 .
Explanation
Analyze the problem and domain We are given the equation 3 x − 1 = 2 x − 5 and asked to find the number of real solutions. First, we need to consider the domain of the square root function. The expression inside the square root must be non-negative, so 3 x − 1 ≥ 0 , which means x ≥ 3 1 . Also, since the square root is non-negative, we must have 2 x − 5 ≥ 0 , so x ≥ 2 5 . Therefore, any solution must satisfy x ≥ 2 5 .
Square both sides and simplify To solve the equation, we square both sides to eliminate the square root: ( 3 x − 1 ) 2 = ( 2 x − 5 ) 2 Simplifying, we get 3 x − 1 = 4 x 2 − 20 x + 25 Rearranging the terms, we obtain a quadratic equation: 4 x 2 − 23 x + 26 = 0
Solve the quadratic equation Now we solve the quadratic equation 4 x 2 − 23 x + 26 = 0 using the quadratic formula: x = 2 a − b ± b 2 − 4 a c where a = 4 , b = − 23 , and c = 26 . The discriminant is D = b 2 − 4 a c = ( − 23 ) 2 − 4 ( 4 ) ( 26 ) = 529 − 416 = 113 Since the discriminant is positive, there are two real solutions for x : x 1 = 8 23 − 113 ≈ 1.546 x 2 = 8 23 + 113 ≈ 4.204
Check for extraneous solutions We need to check if these solutions satisfy the condition x ≥ 2 5 = 2.5 . The first solution, x 1 ≈ 1.546 , does not satisfy this condition, so it is an extraneous solution. The second solution, x 2 ≈ 4.204 , does satisfy this condition. We can verify that x 2 is indeed a solution by plugging it back into the original equation: 3 ( 4.204 ) − 1 ≈ 11.612 ≈ 3.408 2 ( 4.204 ) − 5 ≈ 8.408 − 5 ≈ 3.408 Since the values are approximately equal, x 2 is a valid solution.
Conclusion Therefore, there is only one real solution to the equation 3 x − 1 = 2 x − 5 .
Examples
When designing a bridge, engineers need to calculate the tension in cables using equations that sometimes involve square roots. Understanding how to solve these equations and check for extraneous solutions is crucial to ensure the bridge's stability and safety. Similarly, in physics, calculating the velocity of an object might involve solving an equation with square roots, and extraneous solutions would represent physically impossible scenarios.
