Quadratic Equation

PERSAMAAN KUADRATIK

MTE1014 IPG KAMPUS DARULAMAN PN. HJH SARUVARI BT. BADRU DUJA KHAN

Basics A quadratic equation is written in the Standard Form,

ax  bx  c  0 2

where a, b, and c are real numbers and a  0 . Examples:

x  7 x  12  0 2

x  x  7  0 3 x  4 x  15 2

(standard form)

Basics • 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 Divide by 4 4t = – 1 t=–¼

3t – 5 = 0 3t = 5 t = 5/3

Add 5 Divide by 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 + 6 )(x + 1 )  x2 + 7x + 6 = (x + 6)(x + 1) = 0 Now the equation as given is of the form ab = 0

 set each factor equal to 0 and solve x+6=0 x=–6

x+1=0 x=–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 Add 25  x2 + 10x + 25 = 0

Quadratic equation  factor the left hand side (LHS) x2 + 10x + 25 = (x + 5 )(x + 5 )  x2 + 10x + 25 = (x + 5)(x + 5) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve

x+5=0

x+5=0

x=–5

x=–5

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

Ex: Solve 12y2 – 5y = 2 Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 2  12y2 – 5y – 2 = 0

Quadratic equation  factor the left hand side (LHS) ac method  a = 12 and c = – 2 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) = (3y – 2)(4y + 1)

 12y2 – 5y – 2 = 0  12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve

3y – 2 = 0

4y + 1 = 0

3y = 2

4y = – 1

y = 2/3

y=–¼

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

Ex: Solve 5x2 = 6x Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 6x  5x2 – 6x = 0

Quadratic equation  factor the left hand side (LHS) 5x2 – 6x = x( 5x – 6 )  5x2 – 6x = x(5x – 6) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve 5x – 6 = 0 x=0 5x = 6 x = 6/5 Solution: x = 0 and 6/5  x = {0, 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 = 0 3x2 = 36 x2 = 12 Now take the square root of both sides:

x  12 x   12 2

x   2 2 3

x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 Now take the square root of both sides:

(z  3)  25 2

z  3   25 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 Now take the square root of both sides:

( x  5 )  15 2

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 x2 + 8x + 16 x2 – 6x + 9

Binomial Square (x + 4)2 (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 + 66w = – 4

Add [½(b)]2 to both sides: b = 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

( w  3)  5 2

w 3  5

w  3  5 w  {3  5,3  5}

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

 r2 + (5/2)r (5/2) = (3/2)

Add [½(b)]2 to both sides: b = 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

(r  5 / 4)  49 / 16 2

r  5 / 4  (7 / 4) r  (5 / 4)  (7 / 4) 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 p2 – 6p + 9 = – 7 + 9 (p – 3)2 = 2

 [½(-6)]2 = (-3)2 = 9

(p  3)  2 2

p3  2 p  3 2 p  {3  2,3  2}

The Quadratic Formula • Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero • Completing the square

ax  bx   c b c x  x a a b b c b x  x   a 4a a 4a 2

2

2

2

2

2

2

The Quadratic Formula b b 4ac b x  x   a 4a 4a 4a b  b  4ac  x   2a  4a  2

2

2

2

2

2

2

2

2

Solutions to ax2 + bx + c = 0 for a nonzero are 2

 b  b  4ac x 2a

Ex: Use the Quadratic Formula to solve1x2 + 7x 7 +6=0

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

 b  b 2  4ac x 2a Identify a, b, and c in ax2 + bx + c = 0:

a=1

b= 7

c= 6

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

 7  7 2  4(1)(6) x 2(1)  7  49  24 x 2  7  25 x 2

75 x 2 x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6 x = { - 1, - 6 }

Ex: Use the Quadratic Formula to solve 2 2m2 + 1m – 10 = 0 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by  b  b 2  4ac m 2a

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

a=2

b= 1

c = - 10

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

 1  12  4(2)(10) m 2(2)  1  1  80 m 4  1  81 m 4

1 9 m 4 m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2 m = { 2, - 5/2 }

Any questions . . .