Q. 537 Solve each equation.y23+2y13-3=0... [FREE SOLUTION]

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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 11: Q. 537 (page 1069)

Solve each equation.
$y^{\frac{2}{3}} + 2 y^{\frac{1}{3}} - 3 = 0$

Short Answer

Expert verified

The acquired solutions are:

$y = 1, y = - 27$

But the only solution that will give a real value is:

$y = 1$

Step by step solution

Step 1. Given information.

We have:

$y^{\frac{2}{3}} + 2 y^{\frac{1}{3}} - 3 = 0$

Step 2. Rewrite the equation as standard form

Use the following exponent property: $a^{\frac{n}{m}} = {(a^{\frac{1}{m}})}^{n}$

$y^{\frac{2}{3}} = {(y^{\frac{1}{3}})}^{2} {(y^{\frac{1}{3}})}^{2} + 2 y^{\frac{1}{3}} - 3 = 0 Rewrite 鈥� the 鈥� equation 鈥� with 鈥� y^{\frac{1}{3}} = u u^{2} + 2 u - 3 = 0$

Step 3. Solve the quadratic equation.

We have:

$u^{2} + 2 u - 3 = 0$

For a quadratic equation of the form $a x^{2} + b x + c = 0$ the solutions are:

$x_{1, 鈥� 2} = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a} For 鈥� a = 1, 鈥� b = 2, 鈥� c = - 3 u_{1, 鈥� 2} = \frac{- 2 卤 \sqrt{2^{2} - 4 路鈥� 1 路 (- 3)}}{2 路鈥� 1} u_{1, 鈥� 2} = \frac{- 2 卤鈥� 4}{2 路鈥� 1} u = 1, 鈥� u = - 3$

$Substitute 鈥� back 鈥� u = y^{\frac{1}{3}}, 鈥� solve 鈥� for 鈥� y y^{\frac{1}{3}} = 1 y = 1 y^{\frac{1}{3}} = - 3 y = {(- 3)}^{3} = - 27$

Step 4. Verify the solutions.

At $y = 1 :$

localid="1667913904433" ${(1)}^{\frac{2}{3}} + 2 {(1)}^{\frac{1}{3}} - 3 = 0 1 + 2 - 3 = 0$

$0 = 0$

Hence, verified.

At $y = - 3 :$

${(- 3)}^{\frac{2}{3}} + 2 {(- 3)}^{\frac{1}{3}} - 3 = 0 {(- 1)}^{\frac{2}{3}} 路鈥� 3^{\frac{2}{3}} + 2 {(- 1)}^{\frac{1}{3}} 路鈥� 3^{\frac{1}{3}} - 3 = 0$

Both sides are not equal.

Hence, not verified.

Therefore, localid="1667914096014" $y = 1$ is the real solution.

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91影视

Solve each equation.
$y^{\frac{2}{3}} + 2 y^{\frac{1}{3}} - 3 = 0$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Rewrite the equation as standard form

Step 3. Solve the quadratic equation.

Step 4. Verify the solutions.

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