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