Solve Radical Equation: $\sqrt{-7x+4} = \sqrt{1-x}+3$

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Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of radical equations, specifically tackling this beast: $\sqrt{-7x+4} = \sqrt{1-x}+3$. Now, I know what you're thinking – radicals can look intimidating, like a monster hiding under your math textbook. But don't sweat it! We're going to break this down step-by-step, making it as clear as day. So, grab your favorite snack, settle in, and let's conquer this equation together. Remember, the key to solving any radical equation is to isolate the radical terms and then squaring both sides to get rid of those pesky square roots. We'll also need to be super careful about extraneous solutions, which are solutions that pop up during the solving process but don't actually work in the original equation. It's like finding a cool shortcut that turns out to be a dead end – frustrating, right? But we'll keep an eye out for those! Let's get started by making sure our radicals are defined. For $\sqrt{-7x+4}$ to be a real number, we need $-7x+4 \ge 0$, which means $-7x \ge -4$, or $x \le \frac{4}{7}$. Similarly, for $\sqrt{1-x}$ to be real, we need $1-x \ge 0$, which means $x \le 1$. Combining these, we must have $x \le \frac{4}{7}$ for any real solution to exist. This little check is crucial for spotting those extraneous solutions later on. So, keep this condition in mind as we move forward!

Isolating and Squaring: The First Move

Alright, team, the first major step in solving our radical equation, $\sqrt{-7x+4} = \sqrt{1-x}+3$, is to isolate one of the radical terms. In this case, both radicals are already kind of on their own sides, but the $+3$ on the right is a bit of a nuisance. However, a common strategy is to get one radical by itself. Since $\sqrt{-7x+4}$ is already isolated on the left, we're in a good spot to start squaring. So, let's square both sides of the equation. This is where the magic (and potential trouble!) happens. Squaring the left side is easy peasy: $(\sqrt{-7x+4})^2 = -7x+4$. Now, for the right side, we have $(\sqrt{1-x}+3)^2$. Remember your algebraic identities, guys! This is of the form $(a+b)^2 = a^2 + 2ab + b^2$, where $a = \sqrt{1-x}$ and $b = 3$. So, let's expand it:

$(\sqrt{1-x}+3)^2 = (\sqrt{1-x})^2 + 2(\sqrt{1-x})(3) + (3)^2$

This simplifies to: $1-x + 6\sqrt{1-x} + 9$

Combining the constant terms, we get $10 - x + 6\sqrt{1-x}$.

So, after our first squaring, the equation becomes:

$-7x+4 = 10 - x + 6\sqrt{1-x}$

See? We still have a radical, but it's a bit more manageable. The goal now is to isolate this remaining radical term. Let's move all the non-radical terms to the left side of the equation. We have:

$-7x+4 - (10-x) = 6\sqrt{1-x}$

$-7x+4 - 10 + x = 6\sqrt{1-x}$

Combining like terms on the left: $-6x - 6 = 6\sqrt{1-x}$

Looking good! We can simplify this further by dividing the entire equation by 6:

$-x - 1 = \sqrt{1-x}$

This is a much cleaner equation to work with. We've successfully isolated the radical term. Now, brace yourselves, because it's time for another round of squaring!

The Second Squaring and Solving for x

Okay, we've reached a point where we have $-x - 1 = \sqrt{1-x}$. To eliminate the remaining square root, we need to square both sides again. This is a critical step, but also where extraneous solutions love to sneak in. So, pay extra attention here!

Squaring the left side: $(-x - 1)^2$. Remember that $(-a-b)^2 = (-(a+b))^2 = (a+b)^2$. So, $(-x - 1)^2 = (x+1)^2$. Expanding this gives us $x^2 + 2x + 1$.

Squaring the right side: $(\sqrt{1-x})^2 = 1-x$.

Now, setting the squared sides equal to each other, we get:

$x^2 + 2x + 1 = 1-x$

Awesome! We've successfully eliminated all the radicals. What we have now is a quadratic equation. Our mission is to rearrange this into the standard form $ax^2 + bx + c = 0$ so we can solve for $x$.

Let's move all terms to one side:

$x^2 + 2x + 1 - (1-x) = 0$

$x^2 + 2x + 1 - 1 + x = 0$

Combine like terms:

$x^2 + 3x = 0$

This is a pretty straightforward quadratic equation. We can solve this by factoring. Notice that both terms have an $x$ in common. So, we can factor out $x$:

$x(x + 3) = 0$

For this product to be zero, at least one of the factors must be zero. This gives us two potential solutions:

1. $x = 0$
2. $x + 3 = 0 \implies x = -3$

So, our potential solutions are $x=0$ and $x=-3$. But wait! Remember that warning about extraneous solutions? We must check these values in the original equation to see if they actually work.

Checking for Extraneous Solutions: The Final Verdict

This is arguably the most important step, guys. We found two potential solutions, $x=0$ and $x=-3$, but we need to plug them back into the original equation, $\sqrt{-7x+4} = \sqrt{1-x}+3$, to make sure they hold true. Remember our condition for real solutions: $x \le \frac{4}{7}$. Both $0$ and $-3$ satisfy this, which is a good sign, but it's not a guarantee.

Let's check $x=0$:

Substitute $x=0$ into the original equation:

$\sqrt{-7(0)+4} = \sqrt{1-0}+3$

$\sqrt{0+4} = \sqrt{1}+3$

$\sqrt{4} = 1+3$

$2 = 4$

Uh oh! $2=4$ is false. This means $x=0$ is an extraneous solution. It appeared during our squaring steps but doesn't satisfy the original equation. So, we discard it. Don't get discouraged; this is a normal part of solving radical equations!

Now, let's check $x=-3$:

Substitute $x=-3$ into the original equation:

$\sqrt{-7(-3)+4} = \sqrt{1-(-3)}+3$

$\sqrt{21+4} = \sqrt{1+3}+3$

$\sqrt{25} = \sqrt{4}+3$

$5 = 2+3$

$5 = 5$

Yes! This is true. This means $x=-3$ is a valid solution to the original radical equation.

So, after all that hard work, the only solution that works is $x=-3$. It's a great feeling when you nail it, right? This whole process highlights why checking your solutions is absolutely non-negotiable when dealing with radical equations. Squaring both sides can introduce solutions that aren't actually part of the original problem's set. Keep this in mind for your next math challenge!

Recap and Key Takeaways

Alright, let's quickly recap what we did to solve the radical equation $\sqrt{-7x+4} = \sqrt{1-x}+3$. We started by understanding the importance of the domain for radical equations, ensuring that the expressions under the square roots are non-negative, which gave us $x \le \frac{4}{7}$. Then, we tackled the equation by isolating one radical and squaring both sides. This led to a new equation with one radical still present. We repeated the process: isolate the remaining radical and square both sides again. This transformed the problem into a standard quadratic equation: $x^2 + 3x = 0$. We solved this quadratic by factoring, yielding two potential solutions: $x=0$ and $x=-3$. The final, and most critical, step was to verify these solutions by plugging them back into the original equation. We discovered that $x=0$ was an extraneous solution because it resulted in a false statement ($2=4$), while $x=-3$ was a valid solution as it resulted in a true statement ($5=5$).

So, the key takeaways are:

1. Domain Check: Always determine the domain of the variable to ensure real solutions exist and to help identify potential extraneous solutions early.
2. Isolate and Square: The fundamental technique involves isolating radical terms and squaring both sides to eliminate them. Be prepared to do this potentially more than once.
3. Algebraic Expansion: Master expanding expressions like $(\sqrt{a}+b)^2$ carefully using $(a+b)^2 = a^2 + 2ab + b^2$.
4. Quadratic Equations: Be ready to solve the resulting quadratic equations using factoring, the quadratic formula, or completing the square.
5. Verify Solutions: Crucially, substitute all potential solutions back into the original equation to eliminate extraneous solutions. This step cannot be skipped!

Solving radical equations can be a bit of a journey, involving careful algebra and critical checking. But with practice, you'll get the hang of it! Remember, every complex problem is just a series of smaller, manageable steps. Keep practicing, stay curious, and you'll be a radical equation master in no time. That's all for today, folks! See you in the next article!