Pre Calculus

Question

Solve the equation in the real number system: 9x³+9x²-4x-4=0

a)-3, 0, 3

b)-2/3, 2/3, 1

c)-1, -2/3, 2/3

d)-5, 1, 5

 

Answer 1

9x³ + 9x² - 4x 4 = 0 => (9x³ - 4x) + (9x² - 4) = 0

 

                                 => x * (9x² - 4) + (9x² - 4) = 0

 

                                => (x + 1) * (9x² - 4) = 0

 

∴ x + 1 =0 or 9x²- 4= 0 => x = -1 or x = 2/3 or x = -2/3

Answer 2

9x³+9x²-4x-4 = 0

The real roots must be contained in the factors of 

-4 (the constant) divided by the fators of 9 (the coefficient of the highest powered term.

 

±(1,2,4)

---------- 

±(1,3,9)

 

= ±(1, 1/3, 1/9, 2, 2/3, 2/9, 4, 4/3, 4/9)

The real roots must be contained in this list

of 18 possibilities.

Use synthetic division to test.

 

If 1 is a root then (x-1) is a factor

 

1 | 9   9  -4   -4

   |     9   18  14

    ------------------

     9  18  14   10

Not a factor since the remainder is 10

 

If -1 is a root then (x+1) is a factor

 

-1 | 9   9    -4   -4

    |     -9     0    4

     -------------------

      9   0    -4    0

 

Since the remainder is 0, this is a factor.

9x³+9x²-4x-4 = (x+1)(9x²+0x-4)

 

= (x+1)(9x²-4)

 

Now we need to factor 9x²-4

This is the difference of two squares.

9x²-4 = (3x)²-(2)²

Remember a²-b² = (a-b)(a+b)

9x²-4 = (3x-2)(3x+2)

 

9x³+9x²-4x-4 = (x+1)(3x-2)(3x+2) = 0

 

Roots are:

x+ 1=0 ... x = -1

3x-2=0 ... x = 2/3

3x+2=0 ... x = -2/3

               

The answer is c

 

Answer 3

All you need to do is plug in each answer choice into the equation to check and see if it is valid.  

 

Answer is C