Question

Please help me with these Square Root problems. Part 2. Please Show and check the work.

Answer 1

Answer:

5. 25/36

6. 25/36

7. -3/4 or 6

8. 0 or 4

Step-by-step explanation:

5. Square both sides: x + 4 = 9 - 6$\sqrt{x}$ + x

Subtract x from both sides and subtract 9 from both sides:

-5 = -6$\sqrt{x}$

Square both sides again: 25 = 36x

Divide both sides by 36: x = 25/36

6. Square both sides: x + 4 = 9 + 6$\sqrt{x}$ + x

Subtract x and 9 from both sides: -5 = 6$\sqrt{x}$

Square both sides: 25 = 36x

Divide both sides by 36: x = 25/36

7. Square both sides: x + 3 = 5x + 6 - 6$\sqrt{5x+6}$ + 9

Add 6$\sqrt{5x+6}$ to both sides and isolate it: 6$\sqrt{5x+6}$ = 4x + 12

Divide both sides by 2 and then square both sides again:

3$\sqrt{5x+6}$ = 2x + 6

9 * (5x + 6) = 4x^2 + 24x + 36

45x + 54 = 4x^2 + 24x + 36

4x^2 - 21x - 18 = 0

Factorize: (4x + 3)(x - 6) = 0  ⇒  x = -3/4 or x = 6

8. Square both sides: 2x + 1 = x^2 - 2x + 1

Move all the terms to one side and combine like terms: x^2 - 4x = 0

Factorize: x(x - 4) = 0  ⇒  x = 0 or x = 4

Hope this helps!

Answer 2

Answer:

5. x = 25/36

6. No real solutions

7. x = 6

8. x = 4

Step-by-step explanation:

5. sqrt(x + 4) = 3 - sqrt(x)

Square both sides

x + 4 = 9 - 6sqrt(x) + x

6sqrt(x) = 5

sqrt(x) = 5/6

x = 25/36

6. sqrt(x + 4) = 3 + sqr(x)

x + 4 = 9 + 6sqrt(x) + x

6sqrt(x) = -5

sqrt(x) = -5/6

Not possible. A + square root can not be negative

7. sqrt(x + 3) + 3 = sqrt(5x + 6)

Square both sides

x + 3 + 6sqrt(x + 3) + 9 = 5x + 6

6sqrt(x + 3) = 4x - 6

3sqrt(x + 3) = 2x - 3

Square both sides

9(x + 3) = 4x² - 12x + 9

4x² - 21x - 18 = 0

Using quadratic formula:

x = [21 +/- sqrt(21² - 4(4)(-18)]/(2×4)

x = [21 +/- 27]/8

x = 6, -¾

x = -¾ is rejected because it doesn't satisfy the initial equation

8. sqrt(2x + 1) = x - 1

2x + 1 = (x - 1)²

2x + 1 = x² - 2x + 1

x² - 4x = 0

x(x - 4) = 0

x = 0, 4

0 is rejected because it doesn't satisfy the initial equation