How do you solve the equation and identify any extraneous solutions for #sqrt(x^2 + 2) = x + 4#?

Question

How do you solve the equation and identify any extraneous solutions for #sqrt(x^2 +  2) = x + 4#?

Avery Alexander · Accepted Answer

I found #x=-7/4# I would square both sides to get: #x^2+2=(x+4)^2# #cancel(x^2)+2=cancel(x^2)+8x+16# #8x=-14# #x=-14/8# #x=-7/4#

Eli Assouline · Answer

Square both sides to get:

#x^2+2 = x^2+8x+16#

Hence #x = -7/4# Try squaring both sides to get: #x^2+2 = (x+4)^2 = x^2+8x+16# Subtract #x^2+16# from both sides to get: #8x = -14# Divide both sides by #8# to get: #x = -14/8 = -7/4# Check: #sqrt(x^2+2) = sqrt(49/16+32/16) = sqrt(81/16) = 9/4# #x+4 = -7/4 + 4 = -7/4 + 16/4 = 9/4# graph{(y - sqrt(x^2+2))*(y - x - 4) = 0 [-9.42, 10.58, -1.8, 8.2]}

Isaiah Anthony · Answer

undefined To solve the equation sqrt(x^2 + 2) = x + 4 and identify any extraneous solutions, follow these steps:

1. Square both sides of the equation to eliminate the square root: (sqrt(x^2 + 2))^2 = (x + 4)^2.

2. Simplify the equation: x^2 + 2 = x^2 + 8x + 16.

3. Subtract x^2 from both sides to get rid of the x^2 terms: 2 = 8x + 16.

4. Subtract 16 from both sides to isolate the variable term: -14 = 8x.

5. Divide both sides by 8 to solve for x: x = -14/8 = -7/4.

6. Substitute the found value of x back into the original equation to check for extraneous solutions.

7. sqrt((-7/4)^2 + 2) = -7/4 + 4 simplifies to sqrt(49/16 + 2) = -7/4 + 4.

8. Simplify further: sqrt(49/16 + 32/16) = -7/4 + 64/16.

9. Combine the fractions: sqrt(81/16) = 57/16.

10. Simplify the square root: 9/4 = 57/16.

11. Cross-multiply and solve for x: 9 * 16 = 4 * 57.

12. Simplify: 144 = 228.

13. Since 144 does not equal 228, the solution x = -7/4 is extraneous.

Therefore, the equation sqrt(x^2 + 2) = x + 4 has no valid solutions.