anonymous
  • anonymous
The given equation involves a power of the variable. Find all real solutions of the equation. (If there is no real solution, enter NO REAL SOLUTION.) 3(x − 9)^2/3 = 48
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DebbieG
  • DebbieG
You'll first need to isolate the ( )^a part on the left, by dividing both sides by the coefficient of 3. Then take each side of the equation to the reciprocal of the power, simplify on the right, and solve for x. I'll do a similar example for you, then see if you can do yours :) \(\Large 5(x + 6)^{2/3} = 45\) divide both sides by 5 \(\Large (x + 6)^{2/3} = 9\) take both sides to the reciprocal power (reciprocal of 2/3 is 3/2) \(\Large \left((x + 6)^{2/3}\right)^{3/2} = 9^{3/2}\) Now on the left side, the powers multiply - and since they are reciprocals, you just get a power of 1! (That's the beauty of multiplying by the reciprocal power, see? :) \(\Large x +6= 9^{3/2}\) Now we have to simplify that expression on the right... the thing to remember with rational exponents is, when you can, always take the ROOT first (that's what's in the den'r) and THEN the power: \(\Large 9^{3/2}=\sqrt{9}^3=3^3=27\) make sense? so I have: \(\Large x +6= 27\) which, by subtracting 6 from both sides, gives me: \(\Large x =21\)
DebbieG
  • DebbieG
Of course, it's ALWAYS a good idea to check your work when you solve an equation, by just plugging your solution back into the original equation and making sure that it's true: \(\Large 5(21+ 6)^{2/3} = 5(29)^{2/3}=5\cdot \sqrt[3]{29}^2=5\cdot 3^2=5\cdot 9=45\) So it checks out. :)
anonymous
  • anonymous
I am confused on getting the square room at the right side

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anonymous
  • anonymous
I got 73 as the answer, however my online assignment requires two answers.
DebbieG
  • DebbieG
I was in the midst of replying to this when the site went down earlier... lol. My bad... I forgot one nuance... when the POWER in the original equation is EVEN, then in the step where you take each power to the reciprocal power, you have to use a \(\pm\) on the opposite side. Just like when you solve: \(\Large x^2=9\) and you take the sq root of both sides, you get: \(\Large x=\pm 3\) So to go back to my example, I should have had: \(\Large x + 6 = \pm 9^{3/2}\) which gives me two compound equations: \(\Large x + 6 = 27\) or \(\Large x + 6 =-27\) so: \(\Large x = 21\) or \(\Large x =-33\) And that -33 solution will also check out, in my original equation. Your solution of 73 is correct, btw... just get the solution from the negative sq. root and you should be all set! :)

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