Practice Solving Quadratic Equations |

**INSTRUCTIONS:**

- In all questions except #1, write down the question and solution.

For #1, just write the solutions. - Show all your work and clearly indicate the solutions.
- The "
*nature*" of the solutions is either; when*Real**b² – 4ac*m 0

or(*Complex*); when*Imaginary**b² – 4ac*< 0 - In question #4, for each part, write the quadratic formula,

say it out loud, then substitute the values for*a, b*and*c*into it to solve. - Check your work with the solutions provided. Don't just check answers!

Compare your steps and format to the solutions. Make your work look like mine..

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**QUESTIONS**.

1) Write the solutions by inspection.

a) (

*x – 7*) (*x + 3*) = 0b) (2

*x – 4*) (*x + 6*) = 0c) (

*5x – 3*) (*2x + 7*) = 0d) (

*4x – 2*) ( 9*x + 3*) = 0.

2) Solve these quadratic equations by factoring.

a)

*x² + 2x – 15 = 0*b)

*5x² + 10x = 0*c)

*2x² + 3x – 5 = 0*d)

*10x² – x – 3 = 0*.

3) Solve by completing the square. State the number and nature of the solutions.

a)

*x² – 6x = – 3*b)

*y² – 2y = 2*c)

*2 x² = 3 – x*.

4) Use the quadratic formula to solve these. State the number and nature of the solutions.

a)

*x² + 5x = 3*b)

*9 x² – 6x + 1 = 0*c)

*3 x² – 2x = – 1*.

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**SOLUTIONS**.

1) Write the solutions by inspection.

a) ( *x – 7*) (*x + 3*) = 0*x = 7**or**x = – 3*b) ( *2x – 4*) (*x + 6*) = 0*x = 2**or**x = – 6*c) ( *5x – 3*) (*2x + 7*) = 0*x = 3/5**or**x = – 7/2*d) ( *4x – 2*) ( 9*x + 3*) = 0*x = ½**or**x = – 1/3*.

2) Solve these quadratic equations by factoring.

*x² + 2x – 15 = 0*( *x + 5*) (*x – 3*) = 0*x = – 5**x = 3**5x² + 10x = 0**5x*(*x + 2*) = 0*x = – 2**x = 0**2x² + 3x – 5 = 0*(2 *x + 5*) (*x – 1*) = 0*x = – 5/2**x = 1**10x² – x – 3 = 0*( *2x + 1*) ( 5*x – 3*) = 0*x = – ½**x = 3/5*.

3) Solve by completing the square. State the number and nature of the solutions.

a)

*x² – 6x + 3*= 0 becomes*x² – 6x + 9 – 9 + 3*= 0 becomes (*x – 3*)*² – 6*= 0(

*x – 3*)*² = 6*becomes

There are 2**Real**solutions..

b)

*y² – 2y – 2 = 0*becomes*y² – 2y + 1 – 1 – 2 = 0**y – 1*) ²*– 3 = 0*(

*y – 1*)*² = 3*becomes

There are 2**Real**solutions..

c)

*2 x² + x – 3 = 0*becomes 2(*x² + ½ x*)*– 3*= 0 becomes 2(*x² + ½ x + 1/16*)*– 3*= 02(

*x + ¼*)*² + 1/8 – 24/8 = 0*becomes

There are 2**Real**solutions..

4) Use the quadratic formula to solve these. State the number and nature of the solutions.

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a)

*x² + 5x – 3*= 0*a = 1, b = 5, c = – 3*becomesThere are 2

**Real**roots or solutions..

b)

*9 x² – 6x + 1 =*0*a = 9, b = –6, c = 1*becomesThere is 1

**Real**root or solution. (also called 2 equal roots??).

c) 3

*x*² – 2*x*+ 1 = 0*a*= 3,*b*= – 2,*c*= 1 becomesThere are

**no****Real**roots. There are 2**complex**or**imaginary**roots.

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