Solve $2 \sqrt{3x+4}-9=13$ For X

by Andrew McMorgan 33 views

Hey math whizzes! Ever stare at an equation and think, "What the heck am I supposed to do with this?" Well, you're in the right place, guys. Today, we're diving deep into a juicy algebra problem that involves a square root. We're going to break down how to solve for 'x' in the equation 23x+4βˆ’9=132 \sqrt{3x+4}-9=13. Don't sweat it if radicals make you a bit nervous; we'll tackle this step-by-step, making sure everyone gets it. We've got a few options to choose from – A, B, C, and D – so stick around to see which one is the correct answer! Let's get this algebra party started!

Isolating the Radical: The First Crucial Step

Alright, let's talk about the first big move in solving equations with square roots, like our friend 23x+4βˆ’9=132 \sqrt{3x+4}-9=13. The main goal, my friends, is to get that pesky square root term all by itself on one side of the equals sign. Think of it like giving the radical its own personal space. Right now, it's got a '2' multiplying it and a '-9' hanging out nearby. We need to evict that '-9' first. To do that, we perform the opposite operation. Since it's '-9', we're going to add 9 to both sides of the equation. This is super important – whatever you do to one side, you must do to the other to keep the equation balanced, like a perfectly calibrated scale. So, we add 9 to the left side: 23x+4βˆ’9+92 \sqrt{3x+4}-9+9, which simplifies to 23x+42 \sqrt{3x+4}. And on the right side, we add 9 to 13, giving us 13+9=2213+9=22. Now our equation looks a whole lot friendlier: 23x+4=222 \sqrt{3x+4}=22. See? We're already making progress! This initial step of isolating the radical is absolutely fundamental. If you skip this or mess it up, the rest of the solution will be completely off. So, always remember: isolate first, then deal with the radical itself. It's like peeling an onion; you gotta get through the outer layers before you reach the core.

Dealing with the Coefficient: Making Room for the Root

Now that we've successfully isolated the square root term (hooray!), our equation is 23x+4=222 \sqrt{3x+4}=22. What's the next obstacle? It's that '2' chilling right in front of the square root. This '2' is multiplying our radical. To get the radical truly alone, we need to undo this multiplication. The opposite of multiplying by 2 is dividing by 2, right? You guys know this stuff! So, just like before, we're going to divide both sides of the equation by 2. This keeps everything fair and square (pun intended!). On the left side, 23x+42 \sqrt{3x+4} divided by 2 becomes simply 3x+4\sqrt{3x+4}. And on the right side, we take 22 and divide it by 2, which gives us 11. Now, our equation has transformed into 3x+4=11\sqrt{3x+4}=11. Look at that! The square root term is now completely isolated and ready for the next stage. This step is vital because it sets us up to eliminate the square root entirely. Without dividing by that coefficient, we'd be trying to square something that isn't quite ready yet. So, remember this: after isolating the radical, always divide by any coefficient attached to it. It’s a small step, but it makes a massive difference in simplifying the problem and leading you directly to the solution. You're doing great, team!

Eliminating the Square Root: The Squaring Strategy

We've reached a point where we have 3x+4=11\sqrt{3x+4}=11. This is where the magic of squaring comes in, folks. The square root symbol and the operation of squaring are inverse operations – they cancel each other out. Just like addition and subtraction, or multiplication and division, they undo each other. So, to get rid of the square root on the left side, we need to square it. And to keep our equation balanced, what do we have to do to the right side? You guessed it – square it too! So, we take the entire left side, (3x+4)(\sqrt{3x+4}), and square it. That leaves us with just 3x+43x+4 because (a)2=a(\sqrt{a})^2 = a. Now, we look at the right side: 11211^2. What's 11 times 11? That's 121! So, our equation is now 3x+4=1213x+4=121. This is awesome because we've successfully removed the square root, and we're left with a simple, linear equation. This is the whole point of the previous steps: to get to a place where we can easily solve for 'x'. Always remember that squaring both sides is the key to eliminating a square root. However, a crucial note for your algebra journey: when you square both sides of an equation, you can sometimes introduce extraneous solutions. These are solutions that work in the squared equation but not in the original one. That's why checking your final answer in the original equation is always a good idea. For now, let's keep going with 3x+4=1213x+4=121.

Solving for x: The Final Countdown

We're in the home stretch, guys! Our equation has been simplified down to 3x+4=1213x+4=121. This is a basic two-step equation, and solving for 'x' here is a piece of cake. Our goal is to get 'x' completely by itself. First, we need to get rid of that '+4'. To do this, we subtract 4 from both sides of the equation. On the left, 3x+4βˆ’43x+4-4 simplifies to 3x3x. On the right, 121βˆ’4121-4 equals 117. So now we have 3x=1173x=117. The final step is to undo the multiplication by 3. We do this by dividing both sides by 3. On the left, 3x3x divided by 3 is just xx. On the right, we need to calculate 117Γ·3117 \div 3. Let's break that down: 117=90+27117 = 90 + 27. 90Γ·3=3090 \div 3 = 30, and 27Γ·3=927 \div 3 = 9. So, 117Γ·3=30+9=39117 \div 3 = 30 + 9 = 39. Therefore, x=39x=39. We've found our solution! It's always a good habit to plug this value back into the original equation to ensure it's correct, especially because we squared both sides earlier. Let's check: 23(39)+4βˆ’92 \sqrt{3(39)+4}-9. 3Γ—39=1173 \times 39 = 117. So, 2117+4βˆ’92 \sqrt{117+4}-9. That's 2121βˆ’92 \sqrt{121}-9. We know 121\sqrt{121} is 11. So, 2(11)βˆ’92(11)-9. This equals 22βˆ’922-9, which is 13. Since 13=1313=13, our solution x=39x=39 is correct!

Conclusion: The Answer Revealed

So, after carefully working through each step – isolating the radical, dividing by the coefficient, squaring both sides, and then solving the resulting linear equation – we arrived at our answer. We found that x equals 39. When we checked this solution in the original equation, 23x+4βˆ’9=132 \sqrt{3x+4}-9=13, it held true. This means our value for x is indeed the correct solution. Looking back at the options provided: A. x=0x=0, B. x=26x=26, C. x=2x=2, D. x=39x=39. Our calculated value, 39, matches option D. Great job following along, everyone! Solving equations like these might seem intimidating at first, but by breaking them down into manageable steps and understanding the properties of inverse operations, you can conquer any algebraic challenge. Keep practicing, stay curious, and remember that every problem solved is a win in the world of math. You guys absolutely crushed it!