PLEAS HELP 2. The equation is of the form , where a, b, and c are all positive integers and . Using this equation as a model, create your own equation that has extraneous solutions. (a) Using trial and error with numbers for a, b, and c, create an equation of the form , where a, b, and c are all positive integers and such that 7 is a solution. (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.)
(b) Solve the equation you created in Part 2a.
(c) If your solution in Part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. If your solution in Part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.

Step-by-step explanation:

We have the equation x + 2 = √(3x + 10).

=> (x + 2)² = 3x + 10

=> x² + 4x + 4 = 3x + 10

=> x² + x - 6 = 0

=> (x + 3)(x - 2) = 0

=> x = -3 or x = 2.

However x = -3 is an extraneous solution because x + 2 must be non-negative.

Hence x = 2.

21.4 i did it already for s1

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3