Solve for $x$, where $x$ is a real number.
$\sqrt{8 x-23}-\sqrt{2 x-8}=3$
(If there is more than one solution, separate them with commas.)
Answer (1)
Isolate one square root: 8 x − 23 = 2 x − 8 + 3 .
Square both sides and simplify to get x − 4 = 2 x − 8 .
Square both sides again and simplify to get x 2 − 10 x + 24 = 0 .
Factor the quadratic to find solutions x = 4 and x = 6 . Both solutions are valid. Therefore, 4 , 6 .
Explanation
Understanding the Problem We are given the equation 8 x − 23 − 2 x − 8 = 3 , where x is a real number. We need to solve for x . The expressions inside the square roots must be non-negative: 8 x − 23 ≥ 0 and 2 x − 8 ≥ 0 .
Isolating a Square Root Isolate one of the square roots: 8 x − 23 = 2 x − 8 + 3 .
Squaring Both Sides Square both sides of the equation to eliminate one square root: ( 8 x − 23 ) 2 = ( 2 x − 8 + 3 ) 2 .
Expanding and Simplifying Expand and simplify the equation: 8 x − 23 = ( 2 x − 8 ) + 6 2 x − 8 + 9 . This simplifies to 8 x − 23 = 2 x + 1 + 6 2 x − 8 .
Isolating the Remaining Square Root Isolate the remaining square root: 6 x − 24 = 6 2 x − 8 .
Dividing by 6 Divide both sides by 6: x − 4 = 2 x − 8 .
Squaring Again Square both sides again: ( x − 4 ) 2 = ( 2 x − 8 ) 2 .
Expanding and Simplifying Expand and simplify: x 2 − 8 x + 16 = 2 x − 8 .
Solving the Quadratic Solve the quadratic equation: x 2 − 10 x + 24 = 0 .
Factoring Factor the quadratic: ( x − 6 ) ( x − 4 ) = 0 .
Possible Solutions Find the possible solutions: x = 6 or x = 4 .
Checking Solutions Check the solutions in the original equation: 8 x − 23 − 2 x − 8 = 3 .
If x = 6 : 8 ( 6 ) − 23 − 2 ( 6 ) − 8 = 48 − 23 − 12 − 8 = 25 − 4 = 5 − 2 = 3 . So x = 6 is a solution. If x = 4 : 8 ( 4 ) − 23 − 2 ( 4 ) − 8 = 32 − 23 − 8 − 8 = 9 − 0 = 3 − 0 = 3 . So x = 4 is a solution.
Final Answer Both x = 6 and x = 4 are valid solutions to the equation. Therefore, the solutions are x = 4 , 6 .
Examples
Imagine you're designing a suspension bridge, and the tension in the cables can be modeled by an equation involving square roots, similar to the one we just solved. Finding the correct values for 'x' (representing, say, the length of a support beam) ensures the bridge is stable and safe. Solving such equations helps engineers determine the precise dimensions and forces needed for structural integrity. This blend of algebra and real-world application highlights how mathematical problem-solving is crucial in engineering and design.
