Question

The equation x + 2 = \sqrt{3x + 10}" align="absmiddle" class="latex-formula"> is of the form x + a = $\sqrt{bx + c}$, where a, b, and c are all positive integers and b > 1. Using this equation as a model, create your own equation that has extraneous solutions. (Hint: Substitute 7 for x, and choose a value for a. Then square both sides so you can choose a, b, and c that will make the equation true.) (I know this is a handful but I'm really in need of help right now, thanks to everyone in advance!)

Answer 1

Answer:

Extraneous solution for Equation x + 1 = √7x + 15 : x = - 2

Step-by-step explanation:

We are given the equation x + a = √bx + c. As the hint says, let's substitute 7 for x first, receiving the following equation.

7 + a = √7b + c

Now let's say a = 1 and c = 8. Through substitution we solve for b,

7 + 1 = √7b + 8

8 = √7b + 8

8² = (√7b + 8)²

64 = 7b + 8

7b = 64 - 8 = 56

b = 56 / 7 = 8

Therefore, the equation x + 1 = √8x + 8 ( substitute the value of b ) will have a solution of x = 7. Let's simplify this equation and see if we receive any extraneous solutions.

x + 1 = √8x + 8

(x + 1)² - (√8x + 8)² = 0

(x + 1)² - 8x + 8 = 0

(x + 1)² - 8(x + 1) = 0

(x + 1)(x + 1 - 8) = 0

(x + 1)(x - 7) = 0

x = - 1 and x = 7

We already know that x = 7, but x = - 1 is not an extraneous solution. Therefore let's simplify the equation x + 1 = √7x + 15 and see if we receive any extraneous solutions.

(x + 1)² - (√7x + 15)² = 0

(x + 1)² - 7x + 15 = 0

x² - 5x - 14 = 0

(x - 7)(x + 2) = 0

x = 7 and x = - 2

In this case x = - 2 is an extraneous solution. We can check by substituting it back into the equation x + 1 = √7x + 15.

- 2 + 1 = √7(- 2) + 15

- 1 = √- 14 + 15

- 1 = √1

- 1 ≠ 1 - hence proved that x = - 2 is our extraneous solution