lecture quadratic equations good one

QUADRATIC EQUATIONS

MSJC ~ San Jacinto CampusMath Center Workshop Series

Janice Levasseur

Basics

• A quadratic equation is an equation equivalent to an equation of the type

ax2 + bx + c = 0, where a is nonzero• We can solve a quadratic equation by

factoring and using The Principle of Zero Products

If ab = 0, then either a = 0, b = 0, or both a and b = 0.

Ex: Solve (4t + 1)(3t – 5) = 0

Notice the equation as given is of the form ab = 0

set each factor equal to 0 and solve

4t + 1 = 0 Subtract 1

4t = – 1 Divide by 4

t = – ¼

3t – 5 = 0 Add 5

3t = 5 Divide by 3

t = 5/3

Solution: t = - ¼ and 5/3 t = {- ¼, 5/3}

Ex: Solve x2 + 7x + 6 = 0

Quadratic equation factor the left hand side (LHS)

x2 + 7x + 6 = (x + )(x + )6 1

x2 + 7x + 6 = (x + 6)(x + 1) = 0

Now the equation as given is of the form ab = 0


x + 6 = 0x = – 6

x + 1 = 0x = – 1

Solution: x = - 6 and – 1 x = {-6, -1}

Ex: Solve x2 + 10x = – 25

Quadratic equation but not of the form ax2 + bx + c = 0

x2 + 10x + 25 = (x + )(x + )5 5

x2 + 10x + 25 = (x + 5)(x + 5) = 0



x + 5 = 0

x = – 5

x + 5 = 0

x = – 5

Solution: x = - 5 x = {- 5} repeated root


Add 25 x2 + 10x + 25 = 0

Ex: Solve 12y2 – 5y = 2


ac method a = 12 and c = – 2

= (3y – 2)(4y + 1)


Subtract 2 12y2 – 5y – 2 = 0

ac = (12)(-2) = - 24 factors of – 24 that sum to - 5

1&-24, 2&-12, 3&-8, . . .

12y2 – 5y – 2 = 12y2 + 3y – 8y – 2

= 3y(4y + 1) – 2(4y + 1)

12y2 – 5y – 2 = 0



3y – 2 = 0

3y = 2

4y + 1 = 0

4y = – 1

Solution: y = 2/3 and – ¼ y = {2/3, - ¼ }

y = 2/3 y = – ¼

12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0

Ex: Solve 5x2 = 6x


5x2 – 6x = x( )5x – 6

5x2 – 6x = x(5x – 6) = 0



x = 05x – 6 = 05x = 6

Solution: x = 0 and 6/5 x = {0, 6/5}


Subtract 6x 5x2 – 6x = 0

x = 6/5

Solving by taking square roots

• An alternate method of solving a quadratic equation is using the Principle of Taking the Square Root of Each Side of an Equation

If x2 = a, then x = + a

Ex: Solve by taking square roots 3x2 – 36 = 0

First, isolate x2: 3x2 – 36 = 03x2 = 36

x2 = 12

Now take the square root of both sides:

12x 2 12x

32x x 2 2 3

Ex: Solve by taking square roots 4(z – 3)2 = 100

First, isolate the squared factor:

4(z – 3)2 = 100(z – 3)2 = 25


25)3z( 2

253z z – 3 = + 5

z = 3 + 5

z = 3 + 5 = 8 and z = 3 – 5 = – 2

Ex: Solve by taking square roots 5(x + 5)2 – 75 = 0

First, isolate the squared factor:

5(x + 5)2 = 75(x + 5)2 = 15


2( x 5 ) 15 x 5 15

x 5 15 x 5 15 , x 5 15

Completing the Square

• Recall from factoring that a Perfect-Square Trinomial is the square of a binomial:Perfect square Trinomial Binomial Square x2 + 8x + 16 (x + 4)2

x2 – 6x + 9 (x – 3)2

• The square of half of the coefficient of x equals the constant term: ( ½ * 8 )2 = 16 [½ (-6)]2 = 9

Completing the Square

• Write the equation in the form x2 + bx = c• Add to each side of the equation [½(b)]2

• Factor the perfect-square trinomial x2 + bx + [½(b)] 2 = c + [½(b)]2

• Take the square root of both sides of the equation

• Solve for x

Ex: Solve w2 + 6w + 4 = 0 by completing the square

First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.

w2 + 6w = – 4

Add [½(b)]2 to both sides: b =

6

6 [½(6)]2 = 32 = 9

w2 + 6w + 9 = – 4 + 9

w2 + 6w + 9 = 5

(w + 3)2 = 5

Now take the square root of both sides

5)3w( 2

53w

53w }53,53{w

Ex: Solve 2r2 = 3 – 5r by completing the square

First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.

2r2 + 5r = 3

Add [½(b)]2 to both sides: b =

(5/2)

5/2 [½(5/2)]2 = (5/4)2 = 25/16

r2 + (5/2)r + 25/16 = (3/2) + 25/16

r2 + (5/2)r + 25/16 = 24/16 + 25/16

(r + 5/4)2 = 49/16

Now take the square root of both sides

r2 + (5/2)r = (3/2)

16/49)4/5r( 2

)4/7(4/5r )4/7()4/5(r

r = - (5/4) + (7/4) = 2/4 = ½

and r = - (5/4) - (7/4) = -12/4 = - 3

r = { ½ , - 3}

Ex: Solve 3p – 5 = (p – 1)(p – 2)

Is this a quadratic equation? FOIL the RHS

3p – 5 = p2 – 2p – p + 2

3p – 5 = p2 – 3p + 2

p2 – 6p + 7 = 0

Collect all terms

A-ha . . .

Quadratic Equation complete the square

p2 – 6p = – 7 [½(-6)]2 = (-3)2 = 9

p2 – 6p + 9 = – 7 + 9

(p – 3)2 = 2

2)3p( 2

23p 23p

}23,23{p

The Quadratic Formula

• Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero

• Completing the square 2ax bx c

2b c

x xa a

2 2

2

2 2

b b c bx x

a 4a a 4a

The Quadratic Formula

Solutions to ax2 + bx + c = 0 for a nonzero are

22

2

b b 4acx2a 4a

2b b 4acx

2a

2 2

2

2 2 2

b b 4ac bx x

a 4a 4a 4a

Ex: Use the Quadratic Formula to solve x2 + 7x + 6 = 0

Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by

a2ac4bb

x2

Identify a, b, and c in ax2 + bx + c = 0:

a = b = c = 1

1

7

7

6

6

Now evaluate the quadratic formula at the identified values of a, b, and c

)1(2)6)(1(477

x2

224497

x

2257

x

257

x

x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6

x = { - 1, - 6 }

Ex: Use the Quadratic Formula to solve

2m2 + m – 10 = 0

Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by

a

acbbm

2

42

Identify a, b, and c in am2 + bm + c = 0:

a = b = c = 2

2

1

1

- 10

– 10

Now evaluate the quadratic formula at the identified values of a, b, and c

)2(2)10)(2(411

m2

48011

m

4811

m

491

m

m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2

m = { 2, - 5/2 }

Any questions . . .

lecture quadratic equations good one

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