sq root(2x+50) +2*sq root(x+6) = 5?

Question

sq root(2x+50) +2*sq root(x+6) = 5

help plz show steps

Answer 1

sq root(2x+50) +2*sq root(x+6) = 5
sq root(2x+50) = 5 - 2*sq root(x+6)
Square both sides
2x+50 = 25 - 10 sq root (x+6) + 4(x+6)
2x+50 -25 -4x-24 =-10sq root(x+6)
-2x+1 = -10sq root(x+6)
Square both sides
4x^2 -4x +1 = 100(x+6)
4x2 -104x -599 =0
Use quadratic formula:
x = [104 +/- sqrt(104^2-4*4(-599)]/8
I'll let you finish.

Answer 2

Move the sqrt(2x+50) to the other side by subtracting it

2 sqrt (x+6) = 5 - sqrt(2x+50)

Square both sides (the right side is FOIL)

4 (x+6) = (5 - sqrt(2x+50))(5 - sqrt(2x+50))

4x + 24 = 25 - 10 sqrt (2x+50) + 2x + 50

Move the 10 sqrt (2x+50) to the left and the 4x + 24 to the right by adding the opposite

10 sqrt (2x+50) = -2x + 51

square both sides again

100 (2x+50) = (-2x + 51)(-2x + 51)

Do this out, make it equal to zero, and solve by quadratic formula or if possible FOIL


You have to do it this way because you can't square both sides if things are added. That is you can't just square each thing in the original problem and get 2x+50 + 4(x+6) = 25

Answer 3

example: sq root (x) = (x)61/2

Your Answer:
(2x+50)^1/2 + 2 * (x+6)^1/2 = 5
2x+50 + 2(x+6) = 5^2 = 25
2x+50+2x+12 = 25
4x+62 = 25
4x = 25-62
x = -18.5

Answer 4

Average

Answer 5

by squarin both side
(2x + 50) + 4√[(2x+50)(x+6)] + (x+6) = 25
3x +31 = -4√[(2x+50)(x+6)]
squaring
9x²+ 62x +961 = 16 [2x² +62x+300]
9x² + 62x + 961 = 32x² + 992x +4800
23x² +930x + 3839 = 0
for
ax² + bx + c =0
x = (-b ±√ (b² -4ac))/2a
x = (-930 ±√ (b² -4ac))/2a
x = (-930 ±√ (511712))/46
x = -4.67 or x =-35.77