What is the solution of $4+\sqrt{5 x+66}=x+10$?
A. $x=-10$
B. $x=3$
C. $x=-10$ or $x=3$
D. no solution
Answer (2)
The solution to the equation 4 + 5 x + 66 = x + 10 is x = 3 , after verifying that x = − 10 is an extraneous solution. Therefore, the correct answer is B. x = 3 .
;
Isolate the square root, square both sides, and solve the resulting quadratic equation. Check the solutions in the original equation to eliminate extraneous solutions. The solution to the equation 4 + 5 x + 66 = x + 10 is obtained by:
Isolating the square root: 5 x + 66 = x + 6 .
Squaring both sides and expanding: 5 x + 66 = x 2 + 12 x + 36 .
Rearranging into a quadratic equation: x 2 + 7 x − 30 = 0 .
Factoring the quadratic equation: ( x + 10 ) ( x − 3 ) = 0 , which gives potential solutions x = − 10 and x = 3 .
Checking the solutions: x = − 10 is extraneous, and x = 3 is a valid solution. Therefore, the final answer is 3 .
Explanation
Problem Analysis We are given the equation 4 + 5 x + 66 = x + 10 . Our goal is to find the solution(s) for x and verify them.
Isolating the Square Root First, isolate the square root term by subtracting 4 from both sides: 5 x + 66 = x + 10 − 4 5 x + 66 = x + 6
Squaring Both Sides Next, square both sides of the equation to eliminate the square root: ( 5 x + 66 ) 2 = ( x + 6 ) 2 5 x + 66 = ( x + 6 ) 2
Expanding the Equation Expand the right side of the equation: 5 x + 66 = x 2 + 12 x + 36
Rearranging into Quadratic Form Rearrange the equation into a standard quadratic equation form by subtracting 5 x and 66 from both sides: 0 = x 2 + 12 x + 36 − 5 x − 66 0 = x 2 + 7 x − 30
Solving the Quadratic Equation Now, solve the quadratic equation x 2 + 7 x − 30 = 0 . We can try to factor it:We are looking for two numbers that multiply to -30 and add to 7. These numbers are 10 and -3. So, we can factor the quadratic as: ( x + 10 ) ( x − 3 ) = 0 This gives us two possible solutions: x = − 10 or x = 3 .
Checking for Extraneous Solutions We must check these solutions in the original equation to eliminate any extraneous solutions that may have been introduced when squaring both sides.Let's check x = − 10 : 4 + 5 ( − 10 ) + 66 = − 10 + 10 4 + − 50 + 66 = 0 4 + 16 = 0 4 + 4 = 0 8 = 0 This is false, so x = − 10 is not a solution.Now, let's check x = 3 : 4 + 5 ( 3 ) + 66 = 3 + 10 4 + 15 + 66 = 13 4 + 81 = 13 4 + 9 = 13 13 = 13 This is true, so x = 3 is a solution.
Final Answer Therefore, the only solution to the equation 4 + 5 x + 66 = x + 10 is x = 3 .
Examples
Consider a scenario where you are designing a rectangular garden. The area of the garden is represented by the expression 5 x + 66 , and you want to determine the length x such that adding 4 to the square root of the area equals x + 10 . Solving the equation 4 + 5 x + 66 = x + 10 helps you find the specific value of x that satisfies this condition. This type of problem demonstrates how algebraic equations, including those with square roots, can be used to model and solve practical problems in design and planning.
