What you've done so far is perfectly fine, although it leads you to an incorrect solution. By now you should have experience with equations that have multiple solutions, and this is one of them. Generally speaking, squaring both sides of an equation is okay, but sometimes (particularly when you "clear" a square root), it may either produce extra solutions or cause you to lose solutions. In this case, squaring created an extra "false" solution. Your process of checking it revealed, correctly, that x = 15 is a solution (sometimes they're also called roots or zeroes; they all mean the same thing) of 9(5-3x)=0. However, that's not what we're after here - we want the solutions to 3sqrt(5-3x) = 0. Checking that reveals x = 15 is not a solution:
\(\displaystyle f(x)=3 \cdot \sqrt{5-3x}\)
\(\displaystyle f(15)=3 \cdot \sqrt{5-3(15)}=3 \cdot \sqrt{-40}\)
Oops. That's the square root of a negative number, which can't be done (well, technically it can, but not with the real numbers you're used to working with; you'd need imaginary numbers, which you'll learn about later on). Instead, we need to return to the drawing board and solve the equation a different way to get the actual answer. Starting with the original equation, we can see that, by the
Zero Product Principlehttps://www.mathsisfun.com/algebra/zero-product-property.html, either 3 = 0
or sqrt(5-3x) = 0. We know that 3 = 0 is impossible, so we can just disregard that completely. Or another way of thinking about it is that we can divide both sides of the equation by 3 (0 divided by 3 is still 0). Then we're left with:
\(\displaystyle \sqrt{5-3x}=0\)
From here, we can square both sides without losing any solutions or creating "false" solutions. Try solving that and see what you get. Then check it in the original equation - you should find it checks out.
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