Square root algebra problem? ?

Question

sq. rt 7x +29 =x+3 Find all or real imaginary soluctions to the equation

Answer 1

if the question is √(7x+29)=x+3

Square both sides;

7x+29=x²+6x+9

0=x²-x+20

0=(x+4)(x-5)

x=-4 or x=5

If the question is (√(7x))+29=x+3, then first subtract 29 from both sides before squaring both sides.

_/

Answer 2

im assuming the whole left side is sqrted:

sqrt(7x + 29) = x + 3

square both sides:

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

expand the squared bracket using FOIL or the distributive law:

7x + 29 = x^2 + 3x + 3x + 9

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

take it all to one side:

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

0 = x^2 - x - 20

now, you either factorise, or if this is too difficult, use the quadratic equation to solve for 'x'. in this case, it is easy to factorise:

0 = (x + 4)(x - 5)

therefore, (because one bracket must be zero for the product to be 0) either

x + 4 = 0

x = -4

OR

x - 5 = 0

x = 5

so, the solutions are, (both real) -4 and 5.

subbing them back in, it works:

for x=-4:

sqrt(7*-4 + 29) = -4 + 3

sqrt(-28 + 29)  = -1

sqrt (1)           = -1

1 = -1 ^ 2

1 = 1

LHS = RHS

:)

Answer 3

(x+4) (x-5) The solutions are -4 and 5.

Answer 4

square both sides and solve.