Quadratic equation - completing the square help needed?

Hi,

a) 15 – 6x – 2x² = 0

2x² + 6x = 15

2(x² + 3x + __) = 15
½*3 = 3/2 and (3/2)² = 9/4

2(x² + 3x + 9/4) = 15 + 9/2

2(x + 3/2)² = 39/2

(x + 3/2)² = 39/4
. ________. . . . ____
√(x + 3/2)² = ± √39/4
. . . . . . . . . .__
x + 3/2 = ± √39/2
. . . . . . . __
. . . -3 ± √39
x = ------------- <==ANSWER
. . . . . .2


b) 5x² + 8x – 2 = 0

5x² + 8x = 2

5(x² + 8/5x + __) = 2
½*8/5 = 4/5 and (4/5)² = 16/25

5(x² + 8/5x + 16/25) = 2 + 16/5

5(x + 4/5)² = 26/5

(x + 4/5)² = 26/25
. ________ . . . ._____
√(x + 4/5)² = ± √26/25
. . . . . . . . . __
x + 4/5 = ±√26/5

. . . . . . . .__
. . . - 4 ± √26
x = -------------- <==ANSWER
. . . . . .5

I hope that helps!! :-)

x^2 - 18x + seventy six = 0......upload (18/2)^2 to the two components of the equation. observe 18 is surely the fee of the coefficient of the x term. x^2 - 18x + 80 one + seventy six = 80 one (x - 9)^2 + seventy six = 80 one (x - 9)^2 = 5 x - 9 = 5^(a million/2) or x - 9 = -5^(a million/2) x = 5^(a million/2) + 9 or x = -5^(a million/2) + 9

a) 15 - 6x - 2x² = 0

Rearrange in descending exponential order.
-2x² - 6x + 15 = 0

-2x² - 6x + 15 = 0

Group.
(-2x² - 6x) + 15 = 0

Factor
-2(x² + 3x) + 15 = 0

Add placeholders.
-2(x² + 3x + ___) + 15 + 2(___) = 0

Notice that the second blank is multiplied by 2 to account for what you had to add to complete the square.

Take the coefficient of the x term: 3
Divide it by 2: 3 / 2
Square it: (3 / 2)² = 9 / 4

Add 9 / 4 to both blanks.
-2[x² + 3x + (9 / 4)] + 15 + 2(9 / 4) = 0

x² + 3x + 9 / 4 is the expanded form of a perfect square binomial.

Remember that (a + b)² = a² + 2ab + b². Apply this to what you have.
-2[x² + 3x + (9 / 4)] + 15 + 2(9 / 4) = 0
-2[x + (3 / 2)]² + 15 + 2(9 / 4) = 0

Simplify the rest.
-2[x + (3 / 2)]² + 15 + (9 / 2) = 0
-2[x + (3 / 2)]² + 15(2 / 2) + (9 / 2) = 0
-2[x + (3 / 2)]² + (30 / 2) + (9 / 2) = 0
-2[x + (3 / 2)]² + [(30 + 9) / 2] = 0
-2[x + (3 / 2)]² + (39 / 2) = 0

Subtract (39 / 2) from both sides.
-2[x + (3 / 2)]² + (39 / 2) - (39 / 2) = 0 - (39 / 2)
-2[x + (3 / 2)]² = (-39 / 2)

Divide both sides by -2.
-2[x + (3 / 2)]² / -2 = (-39 / 2) / -2
[x + (3 / 2)]² = (39 / 4)

Take the square root of both sides
√[x + (3 / 2)]² = √(39 / 4)
x + (3 / 2) = ± √(39 / 4)
x + (3 / 2) = ± (√39 / 2)

Subtract (3 / 2) from both sides.
x + (3 / 2) - (3 / 2) = (-3 / 2) ± (√39 / 2)
x = (-3 ± √39) / 2

ANSWER: x = (-3 ± √39) / 2

CHECK USING QUADRATIC FORMULA:

Given: ax² + bx + c = 0
Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Given: -2x² - 6x + 15 = 0
Means: a = -2, b = -6, c = 15

x = [-b ± √(b² - 4ac)] / 2a
x = [-(-6) ± √((-6)² - 4(-2)(15))] / 2(-2)
x = [6 ± √(36 + (120)] / -4
x = [6 ± √(156)] / -4
x = [6 ± √156] / -4
x = [6 ± √(4 * 39)] / -4
x = [6 ± 2√39] / -4
x = [-3 ± √39] / 2
TRUE

b) 5x² + 8x - 2 = 0

5x² + 8x - 2 = 0

Group.
(5x² + 8x) - 2 = 0

Factor
5[x² + (8 / 5)x] - 2

Add placeholders.
5[x² + (8 / 5)x + ___] - 2 - 5(___)

Notice that the second blank is multiplied by -5 to account for what you had to add to complete the square.

Take the coefficient of the x term: (8 / 5)
Divide it by 2: (8 / 5) / 2 = 8 / 10 = 4 / 5
Square it: (4 / 5)² = 16 / 25

Add 16 / 25 to both blanks.
5[x² + (8 / 5)x + (16 / 25)] - 2 - 5(16 / 25) = 0

x² + (8 / 5)x + (16 / 25) is the expanded form of a perfect square binomial.

Remember that (a + b)² = a² + 2ab + b². Apply this to what you have.
5[x² + (8 / 5)x + (16 / 25)] - 2 - 5(16 / 25) = 0
5[x + (4 / 5)]² - 2 - 5(16 / 25) = 0

Simplify the rest.
5[x + (4 / 5)]² - 2 - (16 / 5) = 0
5[x + (4 / 5)]² - 2(5 / 5) - (16 / 5) = 0
5[x + (4 / 5)]² - (10 / 5) - (16 / 5) = 0
5[x + (4 / 5)]² + [(-10 - 16) / 5] = 0
5[x + (4 / 5)]² + (-26 / 5) = 0
5[x + (4 / 5)]² - (26 / 5) = 0

Add (26 / 5) to both sides.
5[x + (4 / 5)]² - (26 / 5) + (26 / 5) = 0 + (26 / 5)
5[x + (4 / 5)]² = (26 / 5)

Divide both sides by 5.
5[x + (4 / 5)]² / 5 = (26 / 5) / 5
[x + (4 / 5)]² = (26 / 25)

Take the square root of both sides
√[x + (4 / 5)]² = √(26 / 25)
x + (4 / 5) = ± √(26 / 25)
x + (4 / 5) = ± (√26 / 5)

Subtract (4 / 5) to both sides.
x + (4 / 5) - (4 / 5) = (-4 / 5) ± (√26 / 5)
x = (-4 ± √26) / 5

ANSWER: x = (-4 ± √26) / 5

CHECK USING QUADRATIC FORMULA:

Given: ax² + bx + c = 0
Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Given: 5x² + 8x - 2 = 0
Means: a = 5, b = 8, c = -2

x = [-b ± √(b² - 4ac)] / 2a
x = [-(8) ± √((8)² - 4(5)(-2))] / 2(5)
x = [-8 ± √(64 + (40)] / 10
x = [-8 ± √(104)] / 10
x = [-8 ± √104] / 10
x = [-8 ± √(4 * 26)] / 10
x = [-8 ± 2√26] / 10
x = [-4 ± √26] / 5
true

a)
15 - 6x - 2x^2 = 0
2x^2 + 6x = 15
(2x^2 + 6x)/2 = 15/2
x^2 + 3x = 15/2
x^2 + 3x/2 + 3x/2 = 15/2
x^2 + 3x/2 + 3x/2 + 9/4 = 15/2 + 9/4
(x^2 + 3x/2) + (3x/2 + 9/4) = 30/4 + 9/4
x(x + 3/2) + 3/2(x + 3/2) = 39/4
(x + 3/2)(x + 3/2) = 39/4
(x + 3/2)^2 = 39/4
x + 3/2 = ±√(39/4)
x = -3/2 ±(√39)/2

b)
5x^2 + 8x - 2 = 0
5x^2 + 8x = 2
(5x^2 + 8x)/5 = 2/5
x^2 + 8x/5 = 2/5
x^2 + 4x/5 + 4x/5 = 2/5
x^2 + 4x/5 + 4x/5 + 16/25 = 2/5
(x^2 + 4x/5) + (4x/5 + 16/25) = 2/5
x(x + 4/5) + 4/5(x + 4/5) = 2/5
(x + 4/5)(x + 4/5) = 2/5
(x + 4/5)^2 = 2/5
x + 4/5 = ±√(2/5)
x = -4/5 ±√(2/5)

Completing the square:
Manipulate the equation into standard quadratic - ax^2+bx+c
in the case of A) -2x^2-6x+15=0
get rid of the c term - -2x^2-6x=-15
get rid of the a term - (-2x^2-6x=-15)/2 = -x^2-3x= -1.5
square half of the b term and add to both sides - -x^2-3x+(-1.5)^2 = -1.5+(-1.5)^2

The resulting equation is factorable into (x-1.5)^2=-5.25^2, and to solve, x-1.5^2 = +/-sqrt 5.25^2

check the algebra but thats the method.

Part a)
2x² + 6x - 15 = 0
x² + 3x = 15/2
x² + 3x + 9/4 = 30/4 + 9/4
(x + 3/2)² = 39/4
x + 3/2 = ±(1/2) √39
x = (-3/2) ±(1/2) √39
x = (-1/2) ( 3 ± √39 )

Part b)
x² + (8/5) x = 2/5
x² + (8/5) x + 16/25 = 26/25
(x + 4/5)² = 26/25
(x + 4/5) = ± √26 / 5
x = - 4 / 5 ± √26/ 5
x = (-1/5) (4 ± √26 )

15 – 6x – 2x² = 0, multiply all by (-1)
2x^2+6x-15=0
D=36-4*2*(-15)=156
x=(-6+/-2√39)/4=(-3+/-√39)/2

5x² + 8x – 2 = 0
D=64-4*5*(-2)=104
x=(-8+/-2√26)/10=(-4+/-√26)/5