Question

Consider the following quadratic equations.

Answer 1

1. x² + 3x + 3 = 0     (not factorable so need to use quadratic formula)

x = $\frac{-3+/-\sqrt{3^{2}-4(1)(3)}} {2(1)}$

x = $\frac{-3+/-\sqrt{9-12}} {2}$

x = $\frac{-3+/-\sqrt{-3}} {2}$

x = $\frac{-3+/-i\sqrt{3}} {2}$

x + $\frac{3+i\sqrt{3}} {2}$ = 0  and   x + $\frac{3-i\sqrt{3}} {2}$ = 0

There are 2 COMPLEX ROOTS

2. x² - 2x - 3 = 0

                 ∧

                1 -3 = -2 (this equals "b" so it is factorable)

  (x - 3)(x + 1) = 0

   x - 3 = 0        x + 1 = 0

      x = 3              x = -1

There are 2 REAL ROOTS

3. x² - 6x + 9 = 0

                   ∧

                 -3 -3 = -6 (this equals "b" so it is factorable)

 (x - 3)² = 0

    x = 3, x = 3

There is 1 REAL DOUBLE ROOT

4. -x² + 3 = 0    

            3 = x²

    +/- √3 = x

There are 2 IRRATIONAL ROOTS

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The quadratic expression −x²+3 has only one real factor, (x−3). FALSE

The solutions to x²−2x−3=0 are x=1 or x=−3   FALSE

The quadratic expression x²+3x+3 has two complex factors, (x + $\frac{3+i\sqrt{3}}{2}$)(x + $\frac{3-i\sqrt{3}} {2}$)  TRUE

The solutions to −x²+3=0 are x=3 or x=−3. FALSE

The quadratic expression x²−6x+9 has two complex factors, (x+3) and (x−3). FALSE

The solutions to x²+3x+3=0 are x = $\frac{-3+i\sqrt{3}} {2}$ or x = $\frac{-3-i\sqrt{3}} {2}$  TRUE

The solution to x²−6x+9=0 is x=3 TRUE

The quadratic expression x²−2x−3 has two real factors, (x−3) and (x+1). TRUE