In Exercises 1–38, solve each radical equation.2(x - 1)¹/³ = (x² ...

Question

In Exercises 1–38, solve each radical equation.2(x - 1)¹/³ = (x² + 2x)¹/³

Accepted Answer

Or the following radical equation solve for B where our equation is three multiplied by B minus two to the one third is equals to B squared plus 12 B to the one third. There are possible answers being B is nine and B is six, B is negative nine, B is six. B is nine B negative six for no solution. Now, our first step is to get rid of our exponent. Our powers are raised to the 1/3 on each set, we can get rid of that by raising everything to the power of three, we will then get three to the third, Which is 27 Multiplied by B -2. On the right side, we'll just get B squared plus 12 B. Now we can try and simplify, We will get 27 B -54. When we distribute our 27 is equals to B squared plus 12 B to solve for B. We want all of our terms on one side equaling zero, You will subtract 27 B and add 54. We then get zero is equals to B squared minus 15 B plus 54. Now, we need to find two terms that will add together to equal negative 15 From part factors of 54. Let's take 54 was p 154 to 3 69. So we need two numbers both negative that will add together to equals negative 15. Well, we can notice six and nine if we both are negative or equals negative 15, 60 equals b squared -6 B -9, B plus flashing A B. In the first one, we get B multiplied By B -6. Factoring A negative nine gives us negative nine multiplied by B minus six, giving us our factors of B - multiplied By B -9. We can set them both equal to zero, B minus six Equals zero, B -9 is equal to zero. OK? B is equal to six. B is equal to nine. Now, we can go ahead and check our solution. B is equal to six and B equals to nine. Let's plug in both numbers back to our personal equation. You'll then get three multiplied by nine minus two to the power of one third. S equals to nine squared plus 12 times nine. I rate the power of 1/3. You get three multiplied by 7 to the 1/3. This then equals to 81 plus one oh eight. Raised the power of one third, get three multiplied by seven to the one third is equals To 1 89 to the 1/3. Now we can rewrite 189 S 27 multiplied by seven. You get three multiplied by 7 to the 1/3. S equals to 27. Slide by seven to the one third, which can be further simplified as three multiplied by seven to the one third as equals to three Multiplied by 7 to the 1/3. So this solution checks out and shape the same with BS equal to six, three, multiplied by six minus two to the one third is equals to six squared Plus 12, multiplied by six To the power of 1/3. You get three multiplied by 4 to the 1/3. It equals to 36 plus 72 To the 1/3. The right side will thin simplify To to the power of 1/3. Finally, we know that one oh eight has factors 27 4. If you look at three multiplied by four to the one third as equals 2, 27 multiplied by four to the one third, Which we can then take the cubit of 27 to get three, multiplied by four to the one third is equals to three multiplied by four to the one third, which checks out we can then confirm the answer to our problem. B is equal to six B is equal to nine which corresponds to answer a. OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 1–38, solve each radical equation.2(x - 1)¹/³ = (x² + 2x)¹/³

Key Concepts

Radical Equations

Cube Roots

Isolating Variables

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