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: $a^{\frac{n}{m}} = {(a^{\frac{1}{m}})}^{n}$

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

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:

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} For a = 1, b = 2, c = - 3 u_{1, 2} = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (- 3)}}{2 \cdot 1} u_{1, 2} = \frac{- 2 \pm 4}{2 \cdot 1} u = 1, u = - 3$

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

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 $y = - 3 :$

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

Both sides are not equal.

Hence, not verified.

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

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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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Solve each equation.y23+2y13-3=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.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Pure Maths

Decision Maths

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Mechanics Maths

Calculus

Study anywhere. Anytime. Across all devices.

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