Chapter 9: Q.186 (page 906)

In the following exercises, solve equations in Quadratic Form for $20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$ .

Short Answer

Expert verified

The solution of the equation is $x = \frac{27}{64}, x = \frac{8}{125}$ .

Step by step solution

Step 1. Given Information.

The given equation is $20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$ .

Step 2. Simplifying .

Simplify $20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$

$20 {(x^{\frac{1}{3}})}^{2} - 23 x^{\frac{1}{3}} + 6 = 0$

Substitute $u = x^{\frac{1}{3}}$ , we get:-

$20 u^{2} - 23 u + 6 = 0$

Step 3. Solving quadratic equation .

Using quadratic formula:-

$20 u^{2} - 23 u + 6 = 0$

20 u^{2} - 23 u + 6 = 0

$u_{1, 2} = \frac{- (- 23) \pm \sqrt{{(- 23)}^{2} - 4 \cdot 20 \cdot 6}}{2 \cdot 20} u_{1, 2} = \frac{- (- 23) \pm 7}{2 \cdot 20} u_{1} = \frac{- (- 23) + 7}{2 \cdot 20}, u_{2} = \frac{- (- 23) - 7}{2 \cdot 20} u = \frac{3}{4}, u = \frac{2}{5}$

Step 4. Simplify

Solve for

$x^{\frac{1}{3}} = \frac{3}{4}; x = \frac{27}{64} x^{\frac{1}{3}} = \frac{2}{5}; x = \frac{8}{125}$

Therefore, The solution of the equation is $x = \frac{27}{64}, x = \frac{8}{125}$ .

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In the following exercises, solve equations in Quadratic Form for $20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$ .

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Simplifying .

Step 3. Solving quadratic equation .

Step 4. Simplify

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Simplifying .

Step 3. Solving quadratic equation .

Step 4. Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

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Study anywhere. Anytime. Across all devices.

In the following exercises, solve equations in Quadratic Form for $20 x^{\frac{2}{3}} - 23 x^{\frac{1}{3}} + 6 = 0$ .