Verified Answer :

  • Identify the domain of the expression as \([-4, \infty)\).
  • Isolate and square the radicals to transform the equation into a quadratic form \(x^2 - 288x + 1136 = 0\).
  • Solve the quadratic equation to find potential candidates \(x = 284\) and \(x = 4\).
  • Verify the candidates against the original equation to exclude extraneous roots, confirming the final result \(\boxed{4}\).

Explanation

  1. Determining the Domain

To solve the equation \(\sqrt{2x+8} + \sqrt{x+5} = 7\), we first identify the domain of validity for the square roots. The expressions under the radicals must be non-negative:

  1. \(2x + 8 \geq 0 \implies 2x \geq -8 \implies x \geq -4\)
  2. \(x + 5 \geq 0 \implies x \geq -5\)

The intersection of these conditions is \(x \geq -4\). Thus, the domain of the equation is \([-4, \infty)\).

  1. Isolating the Radical and Squaring Once

To solve the equation, we isolate one of the radical terms:
\[\sqrt{2x+8} = 7 - \sqrt{x+5}\]
Squaring both sides of the equation yields:
\[(\sqrt{2x+8})^2 = (7 - \sqrt{x+5})^2\]
\[2x + 8 = 49 - 14\sqrt{x+5} + (x+5)\]
Simplifying this expression:
\[2x + 8 = 54 + x - 14\sqrt{x+5}\]
\[x - 46 = -14\sqrt{x+5}\]
Rearranging the terms:
\[46 - x = 14\sqrt{x+5}\]

  1. Squaring the Second Time to Form a Quadratic Equation

Squaring both sides again to eliminate the remaining radical:
\[(46 - x)^2 = (14\sqrt{x+5})^2\]
\[2116 - 92x + x^2 = 196(x + 5)\]
\[2116 - 92x + x^2 = 196x + 980\]
Moving all terms to one side, we obtain the quadratic equation:
\[x^2 - 288x + 1136 = 0\]

  1. Finding and Verifying the Roots

Using the quadratic formula to solve \(x^2 - 288x + 1136 = 0\), we find the roots:
\[x = \frac{-(-288) \pm \sqrt{(-288)^2 - 4(1)(1136)}}{2(1)}\]
\[x = \frac{288 \pm \sqrt{82944 - 4544}}{2}\]
\[x = \frac{288 \pm \sqrt{78400}}{2}\]
\[x = \frac{288 \pm 280}{2}\]
This gives us two potential solutions:

  1. \(x_1 = \frac{288 + 280}{2} = \frac{568}{2} = 284\)
  2. \(x_2 = \frac{288 - 280}{2} = \frac{8}{2} = 4\)

Now we check these values against the domain \(x \geq -4\). Both satisfy the domain. However, we must check if they satisfy the original equation \(\sqrt{2x+8} + \sqrt{x+5} = 7\).
For \(x = 284\): \(\sqrt{2(284)+8} + \sqrt{284+5} = \sqrt{576} + \sqrt{289} = 24 + 17 = 41 eq 7\).
For \(x = 4\): \(\sqrt{2(4)+8} + \sqrt{4+5} = \sqrt{16} + \sqrt{9} = 4 + 3 = 7\).
Therefore, the only valid solution is \(x = 4\).

Examples

In real-life engineering or physics, such as balancing a mechanical beam supported by two springs, you often encounter equations with square roots. These arise from distance formulas or non-linear spring responses. Solving these ensures the beam's equilibrium point is calculated correctly, preventing structural failure. This mathematical approach helps in finding precise design parameters.

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