How do you solve #sqrt(x + 5) = sqrt(x - 3) + 2#?

Question

Kennedy Adams · Accepted Answer

Multiply by the conjugate to eliminate the squares
Use a linear algebra trick to eliminate the #sqrt(x-3)# term.
Square the remaining radical to solve.

Given: #sqrt(x + 5) = sqrt(x - 3) + 2# Subtract #sqrt(x-3)# from both sides and mark it as equation [1]: #sqrt(x + 5) - sqrt(x - 3) = 2" [1]"# To eliminate the radicals on the left, multiply both sides by #sqrt(x + 5) + sqrt(x - 3)#: #(x + 5) -(x - 3) = 2(sqrt(x + 5) + sqrt(x - 3))# #x+ 5 -x + 3 = 2(sqrt(x + 5) + sqrt(x - 3))# #8 = 2(sqrt(x + 5) + sqrt(x - 3))# After dividing each side by two, write equation [2]: #sqrt(x + 5) + sqrt(x - 3) = 4" [2]"# Combine equations [2] and [1] as follows: #2sqrt(x+5) = 6# #sqrt(x+5) = 3 #x+5 = 9# #x = 4# Check: #sqrt(4 + 5) = sqrt(4 - 3) + 2# #sqrt9 = sqrt1+2# #3 = 3# This verifies.

Avery Alexander · Answer

undefined To solve the equation $ \sqrt{x + 5} = \sqrt{x - 3} + 2 $, you can square both sides of the equation to eliminate the square roots. Then, solve for $ x $ by isolating it.

First, square both sides of the equation:

$$ (\sqrt{x + 5})^2 = (\sqrt{x - 3} + 2)^2 $$

$$ x + 5 = (x - 3) + 4\sqrt{x - 3} + 4 $$

Next, simplify the equation:

$$ x + 5 = x - 3 + 4\sqrt{x - 3} + 4 $$

$$ x + 5 = x + 1 + 4\sqrt{x - 3} $$

Now, isolate the square root term:

$$ 4\sqrt{x - 3} = 5 $$

$$ \sqrt{x - 3} = \frac{5}{4} $$

Square both sides again to eliminate the square root:

$$ (\sqrt{x - 3})^2 = \left(\frac{5}{4}\right)^2 $$

$$ x - 3 = \frac{25}{16} $$

Finally, solve for $ x $:

$$ x = \frac{25}{16} + 3 $$

$$ x = \frac{25}{16} + \frac{48}{16} $$

$$ x = \frac{73}{16} $$

So, the solution to the equation is $ x = \frac{73}{16} $.