Intermediate Algebra Help!?

Question

I had my first assignment today and it was out of 40 questions. There's only a select few I don't understand. So any help will be greatly appreciated.

Fraction: 1/2y^2 + 35/12 = 47/12y (should be two solutions)

0.01r - 0.01 = -0.42r^2 (should be two solutions)

(X-9) (X+7) = -28 (should be two solutions)

X^3 + 5x^2 - 9x - 45 = 0 (should be three solutions)

3x^3 + 2x^2 - 108x - 72 =0 (should be three solutions)

Thanks everyone for any help that you give.

Answer 1

I will help you with first one:

1/2y^2 + 35/12 = 47/12y;     multiply everything (both sides) by 12y^2

12y^2 * (1/2y^2 + 35/12 )= 12y^2* (47/12y)

6 + 35y^2 = 47y  or   35y^2 - 47y + 6 = 0

solve by general formula =

x= [-(-47)+/- square root ( (-47)^2 - (4*35*6))] / 2*35   =

[47 +/- square root ( 2209 -840)]  / 70   =

[47 +/- square root (1369)] / 70 =

[47 +/-  37] / 70 ;  

x1 = (47+37)/70 = 84/70 = 12/10 = 6/5

x2 = (47-37)/70 = 10/70 = 1/7

Answer 2

Fraction: 1/2y^2 + 35/12 = 47/12y (should be two solutions)

Is this (1/2)y² + (35/12) = (47/12)y ? If so then:

Multiply all terms by 12 to remove the fractions:

6y² + 35 = 47y

6y² - 47y + 35 = 0

(6y - 5)(y - 7) = 0

6y - 5 = 0 or y - 7 = 0

y = 5/6 or y = 7

0.01r - 0.01 = -0.42r^2 (should be two solutions)

Multiply all terms by 100 to remove the decimals:

1r - 1 = -42r²

42r² + r - 1 = 0

(7r - 1)(6r + 1) = 0

7r - 1 = 0 or 6r + 1 = 0

r = 1/7 or r = -1/6

(X-9) (X+7) = -28 (should be two solutions)

x² + 7x - 9x - 63 = -28

x² - 2x - 63 = -28

x² - 2x - 35 = 0

(x - 7)(x + 5) = 0

x - 7 = 0 or x + 5 = 0

x = 7 or x = -5

X^3 + 5x^2 - 9x - 45 = 0 (should be three solutions)

Factor by grouping:

(x³ + 5x²) - (9x + 45) = 0

x² (x + 5) - 9 (x + 5) = 0

(x + 5)(x² - 9) = 0 [Note: x² - 9 is the difference of two squares.]

(x + 5)(x - 3)(x + 3) = 0

x + 5 = 0 or x - 3 = 0 or x + 3 = 0

x = -5 or x = 3 or x = -3

3x^3 + 2x^2 - 108x - 72 =0 (should be three solutions)

Factor by grouping:

(3x³ + 2x²) - (108x + 72) = 0

x² (3x + 2) - 36 (3x + 2) = 0

(3x + 2)(x² - 36) = 0 [Note: x² - 36 is the difference of two squares.]

(3x + 2)(x - 6)(x + 6) = 0

3x + 2 = 0 or x - 6 = 0 or x + 6 = 0

x = -2/3 or x = 6 or x = -6