Find all real or imaginary solutions to each equation?

Question

*square root*[7x + 29]= x+3

Answer 1

squaring both sides

7x + 29 = x^2 + 6x + 9
x^2 - x - 20 = 0
(x - 5)(x+4) = 0

x = -4 is extraneous

thus x = 5


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Answer 2

Square both sides. You then get:

7x+29 = (x+3)^2

7x+29 = x^2 + 6x + 9

x^2 - x - 20 = 0

You can now solve this using the quadratic formula -- or factor it.

I will leave that part to you.

This gives you two solutions. Because we squared the original equations, both roots may not give you the right answer -- so you need to put them into the original equation and see if you get a correct answer.

Answer 3

sq. the two aspects. then you definitely get: 7x+29 = (x+3)^2 7x+29 = x^2 + 6x + 9 x^2 - x - 20 = 0 you may now sparkling up this utilising the quadratic formulation -- or factor it. i visit go away that section to you. this supplies 2 strategies. by using fact we squared the unique equations, the two roots will possibly no longer provide you the best answer -- so so you might place them into the unique equation and notice in case you get a best answer.

Answer 4

sqrt (7x + 29) = x + 3
solution:
(sqrt (7x + 29) )^2= (x + 3)^2
7x + 29= x^2+6x+9
0=x^2+6x+9 -7x - 29
0=x^2 -x - 20
0=(x -5) (x +4)
x =5 x = -4
i hope it helps.