Solving Sqrt(7x-26) - Sqrt(x-5) = 3 A Step-by-Step Guide

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Hey guys! Today, we are going to tackle a fun little algebraic problem. We need to solve for x in the equation $\sqrt{7x-26}-\sqrt{x-5}=3$, making sure that x is a real number. Now, before we dive in, let's lay out a plan of attack. Our main goal is to isolate x, but we've got these pesky square roots in the way. So, our strategy will involve getting rid of those square roots by squaring both sides of the equation. However, we need to be careful because squaring can sometimes introduce extraneous solutions, which are solutions that pop up during the solving process but don't actually work in the original equation. Therefore, we'll need to check our answers at the end to make sure they're legit.

Step-by-Step Solution

  1. Isolate one of the square roots. It's generally a good idea to start by isolating one of the square root terms. Let's add $\sqrt{x-5}$ to both sides of the equation:

    7x−26=x−5+3\sqrt{7x-26} = \sqrt{x-5} + 3

  2. Square both sides. Now, to get rid of the square root on the left, we'll square both sides of the equation. Remember, we need to square the entire right side, not just the individual terms:

    (7x−26)2=(x−5+3)2(\sqrt{7x-26})^2 = (\sqrt{x-5} + 3)^2

    This simplifies to:

    7x−26=(x−5+3)(x−5+3)7x - 26 = (\sqrt{x-5} + 3)(\sqrt{x-5} + 3)

    Expanding the right side needs care. When expanding the right side, it's important to remember the distributive property (often referred to as FOIL - First, Outer, Inner, Last). We get:

    7x−26=(x−5)+6x−5+97x - 26 = (x - 5) + 6\sqrt{x-5} + 9

    Combining the constants on the right side, we have:

    7x−26=x+4+6x−57x - 26 = x + 4 + 6\sqrt{x-5}

  3. Isolate the remaining square root. We still have a square root term, so let's isolate it. Subtract x and 4 from both sides:

    6x−30=6x−56x - 30 = 6\sqrt{x-5}

  4. Simplify the equation. We can simplify this equation by dividing both sides by 6:

    x−5=x−5x - 5 = \sqrt{x-5}

  5. Square both sides again. To eliminate the remaining square root, we square both sides one more time:

    (x−5)2=(x−5)2(x - 5)^2 = (\sqrt{x-5})^2

    Expanding the left side gives us:

    x2−10x+25=x−5x^2 - 10x + 25 = x - 5

  6. Rearrange into a quadratic equation. Now, we'll rearrange the equation into a standard quadratic form by subtracting x and adding 5 to both sides:

    x2−11x+30=0x^2 - 11x + 30 = 0

  7. Solve the quadratic equation. We can solve this quadratic equation by factoring. We're looking for two numbers that multiply to 30 and add up to -11. Those numbers are -5 and -6. So, we can factor the quadratic as:

    (x−5)(x−6)=0(x - 5)(x - 6) = 0

    This gives us two potential solutions:

    x=5 or x=6x = 5 \text{ or } x = 6

  8. Check for extraneous solutions. This is the crucial step! We need to plug each potential solution back into the original equation to make sure it works. Let's start with x = 5:

    7(5)−26−5−5=35−26−0=9−0=3\sqrt{7(5)-26} - \sqrt{5-5} = \sqrt{35-26} - \sqrt{0} = \sqrt{9} - 0 = 3

    So, x = 5 works!

    Now, let's check x = 6:

    7(6)−26−6−5=42−26−1=16−1=4−1=3\sqrt{7(6)-26} - \sqrt{6-5} = \sqrt{42-26} - \sqrt{1} = \sqrt{16} - 1 = 4 - 1 = 3

    So, x = 6 also works!

Final Answer

Therefore, the solutions to the equation $\sqrt{7x-26}-\sqrt{x-5}=3$ are x = 5 and x = 6.

Key Concepts Recap

Before we wrap up, let's briefly recap the key concepts we used to solve this problem:

  • Isolating square roots: We strategically isolated the square root terms to eliminate them by squaring.
  • Squaring both sides: Squaring both sides of the equation is a powerful technique for getting rid of square roots, but it's essential to remember to square the entire side, not just individual terms.
  • Expanding expressions: Careful expansion of expressions, especially when squaring binomials involving square roots, is crucial for avoiding errors.
  • Solving quadratic equations: We encountered a quadratic equation, which we solved by factoring. Other methods, such as the quadratic formula, could also be used.
  • Checking for extraneous solutions: This is the most important step when dealing with equations involving square roots. Squaring can introduce solutions that don't actually satisfy the original equation, so we must always check our answers.

Why Checking for Extraneous Solutions is So Important

Guys, I can't stress this enough: checking for extraneous solutions is absolutely essential when you're dealing with radical equations (equations with square roots, cube roots, etc.). The reason is that the squaring operation can sometimes make a false statement true. Let me give you a simple example:

Suppose we have the equation $x = -2$. This equation has one solution: x = -2. Now, let's square both sides:

x2=(−2)2x^2 = (-2)^2

x2=4x^2 = 4

Now, we have a new equation, $x^2 = 4$. This equation has two solutions: x = 2 and x = -2. Notice that we've gained a solution (x = 2) that wasn't there in the original equation. This is an extraneous solution.

The same thing can happen in more complex equations involving square roots. When we square both sides, we're essentially creating a new equation that might have more solutions than the original one. That's why we need to plug our solutions back into the original equation to see if they actually work.

In our problem, both x = 5 and x = 6 happened to be valid solutions, but it's very possible that one of them (or both!) could have been extraneous. If we had skipped the checking step, we might have included an incorrect answer in our final result.

Tips for Avoiding Mistakes

Alright, so how can we minimize the chances of making mistakes when solving radical equations? Here are a few tips:

  • Be careful when squaring binomials: Remember that $(a + b)^2 = a^2 + 2ab + b^2$, not $a^2 + b^2$. This is a very common error.
  • Isolate the square root before squaring: This will make the algebra much cleaner.
  • Simplify as you go: If you see an opportunity to simplify the equation, take it! This will reduce the risk of making errors later on.
  • Write neatly and organize your work: This might seem like a small thing, but it can make a big difference. If your work is messy and disorganized, it's much easier to make a mistake.
  • Double-check your work: Before you declare victory, go back and review your steps. Did you make any algebraic errors? Did you forget to check for extraneous solutions?

Alternative Approaches

While our approach of isolating and squaring is a standard method, there might be alternative ways to solve this equation. For example, we could have tried a substitution. Let's say we let $y = \sqrt{x-5}$. Then, $y^2 = x - 5$, so $x = y^2 + 5$. We can substitute this into the original equation:

7(y2+5)−26−y=3\sqrt{7(y^2 + 5) - 26} - y = 3

7y2+35−26−y=3\sqrt{7y^2 + 35 - 26} - y = 3

7y2+9=y+3\sqrt{7y^2 + 9} = y + 3

Now we can square both sides and solve for y. Once we find the values of y, we can substitute back to find x. This approach might seem a bit more complicated, but it's a good illustration of how there can be multiple paths to the same solution.

Conclusion

Solving equations with square roots can be a bit tricky, but with careful attention to detail and a systematic approach, you can conquer them! Remember to isolate the square roots, square both sides (and expand carefully!), simplify, and always check for extraneous solutions. And hey, don't be afraid to explore different solution methods – sometimes a fresh perspective can make all the difference.

I hope this breakdown has been helpful, guys. Keep practicing, and you'll become square root-solving pros in no time!