square root of (7x + 29) = x + 3?

Question

find all real or imaginary soulutions to this equation. use the method of your choice.

Answer 1

7x + 9 = (x + 3)²
7x + 9 = x² + 6x + 9
x² - x = 0
x (x - 1) = 0
x = 0 , x = 1

Answer 2

right this is the unique question: ?(7x+29) = x+3 to get rid of the basis sign, sq. the two facets: 7x + 29 = (x+3)(x+3) Multiply by way of: 7x + 29 = x² + 6x + 9 team all of it on one element: x² -x -20 = 0 ingredient it: (x - 5)(x + 4) = 0 x = 5 or x = -4 Double verify the solutions by skill of plugging on your solutions. x = 5 ?(7(5) + 29) =? (5) + 3 ?(35 + 29) =? 8 ?sixty 4 =? 8 8 = 8 <-- verify, so x = 5 is right x = -4 ?(7(-4) + 29) =? (-4) + 3 ?(-28 + 29) =? -a million ?a million =? -a million (?a million = ±a million) <-- verify, so x = -4 is right The solutions for x are: x = 5 x = -4

Answer 3

Ok, here's what I come up with:

(7X + 29)^(1/2) = X + 3
square both sides

7X + 29 = (X+3) (X+3)
then,
29 = X^2 + 6X - 7X + 9

29 = X^2 -X + 9, therefore:

0 = X^2 - X + 9 -29

0 = X^2 - X - 20

thus, (X - 5) (X + 4) = 0

The roots are 5 and -4

Check your answer by plugging the roots back into the original equation and seeing if the answer is true.

Answer 4

You should really do your own homework.

However, 'cause I'm a nice guy, I'll help you get started.

You need to solve for x. This means x needs to be by itself on one side of the equals sign.

Remember that anything you do to one side of the equation must be done to the other side as well.

The first thing you'll want to do is is square both sides to get rid of that nasty square root business.
This will give you
(7x + 29) = (x + 3) squared or (x+3)(x+3)

use FOIL (First Outside Inside Last) to multiply out (x+3)(x+3)
Subtract what you have on the right side of the = sign from both sides of the equation to get an equation of the form:
ax^2 + bx + c = 0
Then apply the quadratic equation to this formula to find all possible values for x.

Answer 5

7x + 29 = x^2 + 6x + 9
x^2 -x -20=0
x^2 -5x + 4x -20 = 0
so x = 5 and -4