The equation x + 2 = [tex]\sqrt{3x + 10}[/tex] is of the form x + a = [tex]\sqrt{bx + c}[/tex], 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!)

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

MrRoyal MrRoyal · Answer 2 · 2021-12-30T18:16:26+01:00

The equation with extraneous solution is [tex]x +5 = \sqrt{6x + 4[/tex]

The equation is given as:

[tex]x + 2 = \sqrt{3x + 10[/tex]

The form of the equation is given as:

[tex]x +a = \sqrt{bx + c[/tex]

Now, we assume that x = 7.

So, the equation becomes

[tex]7 +a = \sqrt{7b + c[/tex]

Assume that the value of a is 5

[tex]7 +5 = \sqrt{7b + c[/tex]

Add 7 and 5

[tex]12 = \sqrt{7b + c[/tex]

Square both sides of the equation

[tex]144= 7b + c[/tex]

Assume that the value of b is 6.

So, the equation becomes

[tex]144= 7(6) + c[/tex]

[tex]144= 42 + c[/tex]

Assume that c is 4

[tex]144= 42 + 4[/tex]

[tex]144= 46[/tex]

The above equation is false because 144 does not equal 46.

The values of a, b and c that makes the equation false are:

[tex]a=5[/tex]

[tex]b = 6[/tex]

[tex]c = 4[/tex]

So, we have:

[tex]x +a = \sqrt{bx + c[/tex]

[tex]x +5 = \sqrt{6x + 4[/tex]

Hence, the equation with extraneous solution is [tex]x +5 = \sqrt{6x + 4[/tex]

Read more about extraneous solutions at:

https://brainly.com/question/4563555