Chapter 9: Q. 423 (page 1108)

In the following exercises, solve.
$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Short Answer

Expert verified

Part (a) : $\frac{2401}{100}$

Part (b) : 6

Step by step solution

Step 1. Given Information

The given expressions are

$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Step 2. Squaring on both sides

$(a) \sqrt{x} + 10 = \sqrt{x + 2} {(\sqrt{x} + 10)}^{2} = {(\sqrt{x + 2})}^{2} 98 + 20 \sqrt{x} = 0 20 \sqrt{x} = - 98 S q u a r i n g o n b o t h s i d e s {(20 \sqrt{x})}^{2} = 9604 x = \frac{2401}{100}$

The solution is $\frac{2401}{100}$

$(b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4} {(\sqrt{y - 2} + 2)}^{2} = {(\sqrt{2 y + 4})}^{2} - 2 + 4 \sqrt{y - 2} = y 4 \sqrt{y - 2} = y + 2 s q u a r i n g o n b o t h s i d e s 16 (y - 2) = y^{2} + 4 + 4 y 16 y - 32 = y^{2} + 4 + 4 y y^{2} - 12 y + 36 = 0 y = 6$

The solution is 6

Step 3. The solution

The solution for are $\frac{2401}{100} a n d 6$

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In the following exercises, solve.
$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Squaring on both sides

Step 3. The solution

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In the following exercises, solve.ax+10=x+2by-2+2=2y+4

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Squaring on both sides

Step 3. The solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Applied Mathematics

Statistics

Theoretical and Mathematical Physics

Probability and Statistics

Geometry

Study anywhere. Anytime. Across all devices.

In the following exercises, solve.
$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$