another lost math student needs your help!!!?

Question

Please help me solve the following:

(1) sqrt(4x +1) + 3 = 0

(2) sqrt(48 - 2y) + y

(3) sqrt(2y + 7) + 4 = y

(4) sqrt(x+4) = 3

(5) Divide, rationalize the denominator if necessary, then simplify the radical expression:
(6 + sqrt20)/2

Answer 1

The answer to number 5:

3+sqrt5

You know, Yahoo! Answers should not be a place where kids can put their homework on it for other people to do. That's called laziness and will get you no where in life

Answer 2

1. sqrt(4x + 1) + 3 = 0
sqrt(4x + 1) = - 3 square both sides
4x + 1 = 9
4x = 8
x = 2
2. sqrt(48 - 2y) + y = 0
sqrt(48 - 2y) = -y square both sides
48 - 2y = y^2
y^2 + 2y - 48 = 0 use quadratic formula
y = 6, -8
(3) sqrt(2y + 7) + 4 = y
sqrt(2y +7) = y - 4 square both sides
2y + 7 = y^2 - 8y + 16
y^2 - 10y + 9 = 0 use the quadratic formula
y = 9, 1
(4) sqrt(x+4) = 3 square both sides
x + 4 = 9
x = 5
(5) Divide, rationalize the denominator if necessary, then simplify the radical expression:
(6 + sqrt20)/2
(6 + 2sqrt10)/2
3 + sqrt10
ANSWER
3 + sqrt 10

Answer 3

the answer to number 1 is not 2.

sub 2 into the equation and you get:

sqrt(4*2 + 1) + 3 = 0

sqrt(8 +1 ) + 3 = 0

sqrt (9) + 3 = 0

3 + 3 does not = 0 it equals 6 ... so 2 is the wrong answer.

you want the sqrt(of something) = -3

the only "of something" to work will be 9i^4 where i is sqrt(-1)

so you want 9i^4 = 4x + 1

or x = (9i^4-1)/4

now i^4 = 1 ... how nice

the problem arises because when you square both sides
you get (4x+1) = (+or-3)^2

when solving for 4x + 1 = (3)^2 ... the answer is x = 2

however for (4x +1) = (-3)^2 its not

i am sure thier is a more elegant way to prove this but its been way to long since i have mucked around with imaginary numbers.

anyways are you sure you typed in the problem for 1 correctly and its not a typo.

Answer 4

(1) sqrt(4x +1) + 3 = 0
sqrt(4x+1)=-3 square both sides
4x+1=9 subtract 1
4x=8 divide by 4
x=2

(2) sqrt(48 - 2y) + y=0
sqrt(48-2y)=-y
48-2y=y^2
y^2+2y-48=0
(y+8)(y-6)=0
y+8=0
y=-8
y-6=0
y=6

(3) sqrt(2y + 7) + 4 = y
sqrt(2y+7)=y-4
2y+7=y^2-8y+16
y^2-10y+9=0
(y-9)(y-1)=0
y-9=0
y=9
y-1=0
y=1

(4) sqrt(x+4) = 3
x+4=9
x=5

(5) Divide, rationalize the denominator if necessary, then simplify the radical expression:
(6 + sqrt20)/2 =3+sqrt(5)

Answer 5

I will help you with the first one and hopefully you can use it to solve the other problems;

the concept here is the same as any "solve for x", to get the x term by itself.

In the problem sqrt(4x+1)+3=0, first get the sqrt by itself by subtracting both sides by 3:

sqrt(4x+1)= -3

Then get rid of the sqrt by squaring both sides:

4x+1= (-3)squared or 9

Then get the X term by itself by subtracting 1 from both sides:

4x= 8

Then get the X by itself by dividing both sides by 4:

x= 2

hope this helps!