can someone show me how to solve these two equations?

Question

http://img.photobucket.com/albums/v60/simple_3/MATH.jpg

Thank You.

Answer 1

1) sqrt(c + 1) + sqrt(c - 2) = sqrt(3c)

Square both sides, to get

[ sqrt(c + 1) ]^2 + 2sqrt(c + 1)sqrt(c + 2) + [ sqrt(c - 2) ]^2 = 3c

(c + 1) + 2sqrt(c + 1)sqrt(c + 2) + (c - 2) = 3c

Group like terms,

2c - 1 + 2sqrt(c + 1)sqrt(c + 2) = 3c

Moving everything but the square roots to the right hand side gives us

2sqrt(c + 1)sqrt(c + 2) = c + 1

We can combine the product of two square roots as one square roots of the product of the arguments.

2sqrt[ (c + 1)(c + 2) ] = c + 1

Square both sides,

4 [ (c + 1)(c + 2) ] = (c + 1)^2

4(c + 1)(c + 2) = c^2 + 2c + 1
4(c^2 + 3c + 2) = c^2 + 2c + 1

4c^2 + 12c + 8 = c^2 + 2c + 1

Moving everything over to the left hand side,

3c^2 + 10c + 7 = 0

Factor,

(3c + 7)(c + 1) = 0

Equate each factor to 0, to get

c = { -7/3, -1 }

However, we should test if any of them are extraneous solutions. In fact, both of them are extraneous, because our equation is

sqrt(c + 1) + sqrt(c - 2) = sqrt(3c)

Which means, if c is a negative number, sqrt(3c) is not a real number. Therefore, there is no solution.

2) 3y = (5y + 6) / (2y + 3)

Multiply both sides by (2y + 3), to obtain

3y(2y + 3) = 5y + 6

Expand the left hand side,

6y^2 + 9y = 5y + 6

Move everything to the left hand side,

6y^2 + 4y - 6 = 0

Divide by 2,

3y^2 + 2y - 3 = 0

Use the quadratic formula.

y = [ -2 +/- sqrt(2^2 - 4(3)(-3)) ] / (2*3)
y = [ -2 +/- sqrt(4 + 36) ] / (6)
y = [ -2 +/- sqrt(40) ] / 6
y = [ -2 +/- 2sqrt(10) ] / 6
y = [ -1 +/- sqrt(10) ] / 3

Answer 2

Puggy has everything rite in the method of how to do this first problem however I believe there was a minor mistake when she squared the radical sum. Putting a sqrt(c+2) instead of sqrt(c-2). So doing essentially the same work I came up with the answers of:

c={-1,3} where c=3 is the only real answer/ solution.

Answer 3

[√(c + 1) + √(c - 2) ] ² = [√(3c) ] ²
(c + 1) + 2√(c - 1) (c + 1) + c - 2 = 3c
2√ [(c + 1)(c - 2) ] = c + 1
4(c + 1) (c - 2) = c² + 2c + 1
4 (c² - c - 2) = c² + 2c + 1
3c² - 6c - 9 = 0
c² - 2c - 3 = 0
(c - 3) (c + 1) = 0
c = 3 , c = - 1
Accept c = 3 as valid answer.

Question 2
3y (2y + 3) = 5y + 6
6y² + 9y = 5y + 6
6y² + 4y - 6 = 0
3y² + 2y - 3 = 0
y = [ - 2 ± √(4 + 36) ] / 6
y = [ - 2 ± √40 ] / 6
y = [ - 2 ± 2√10 ] / 6
y = [ - 1 ± √10 ] / 3

Answer 4

On the first one, square both sides of the equation I believe and remember that squaring the part within parenthesis is equivalent to multiplying it by itself. Use the FOIL method to do the multiplication. Then solve for C.

Answer 5

1. c=-1
2. y=.72 and -1.4