math fustrate me!!!!!!!!! plzzz help?

Question

can you help me with these two problems.... i promise i'll get
~solve the equation by completing the square~
1) p²+5p+6=0
2)2c²+c=5

a word of advie don't be absent from school... ever!!!!

i feel like a blond now (no offense many blonds are smart) all you have to do is get the co-efficient of x divide by 2 and square it, and add it to both sides of the equal side right??? thanx trying to keep my gpa a 4.0......

Answer 1

you want the equations in the format
(x+a)^2 + b


1) p^2 + 5p + 6
Since we have 5, let's put 5/2 in the bracket
(p + 2.5)^2 = p^2 + 5p + 6.25

So to get the original equation we must now subtract 0.25
Answer: (p + 2.5)^2 - 0.25

2) c^2 + c - 5
Let's use 1/2
(c + 0.5)^2 = c^2 + c + 0.25
We must subtract 5.25
Answer: (c + 0.5)^2 - 5.25

Answer 2

Reviewing,
(a + b)² = a² + 2ab +b²
so you need to get to that form
1)
p² + 5p + 6 = 0
p² + 2(5/2)p = - 6 gives us 2ab, so b is 5/2
adding (5/2)² to both sides,
p² + 2(5/2)p + (5/2)² = - 6 + (5/2)²
(p + 5/2)² = - 6 + 25/4 = 1/4
p + 5/2 = ± √(1/4) = ± 1/2
p = - 5/2 - 1/2, - 5/2 + 1/2
p = - 3, - 2

2)
2c² + c = 5
We want a unit coefficient for c, so divide thru by 2
c² + (1/2)c = 5/2
c² + 2(1/4)c + (1/4)² = 5/2 + (1/4)²
(c + 1/4)² = 40/16 + 1/16
c + 1/4 = ± √(41/16)
c = - 1/4 ± (1/4)√41
c = - 1/4 + (1/4)√41, c = - 1/4 - (1/4)√41

Answer 3

To complte the square, we must add (b/2)^2 to each side, where b is the number in front of the x term
p^2 + 5p + 6 = 0====> your b is 5
p^2 + 5p = -6
p^2 + 5p +____ = -6 + _____
(5/2)^2 = 25/4, we will add to each side
p^2 + 5p + 25/4 = -6 + 25/4
(p+5/2)^2 = 1/4 {we complete the square by factoring}
now square root each side, and need +/-
p + 5/2 = +/- sqrt(1/4)
p = -5/2 +/- 1/2
p = -5/2 - 1/2 OR p = -5/2 + 1/2
p= -3 OR p= -2

:)

the next one divide by two to make it easier!
2c^2 + c = 5
c^2 + 1/2 c = 5 ====> your b=1/2===> (b/2)^2 = (1/4)^2 = 1/16
c^2 + 1/2 c + ______ = 5/2 + _______
c^2 + 1/2 c + 1/16 = 5/2 + 1/16
(c+1/4)^2 = 41/16 ===> sqrt each side
c+1/4 = +/- sqrt (41/16)
c = -1/4 +/- (sqrt41)/4
c= -1/4 + (sqrt41)/4 OR c = -1/4 - (sqrt41)/4

:)
good luck!

Answer 4

completing the square is a technique where u add the same number to both sides....if u think about it, when u add the same number to both sides ur not changing the equation at all

1. p²+5p+6=0
p²+5p=-6 (move the constant over)
p²+5p+(25/4)=-6+(25/4) (with the equation on the left, u do (b/2)^2
(p+5/2)²=1/4
p+5/2=+/- 1/2
p = -2, or -3

2. try on ur own =)

Answer 5

1) p²+5p+6=0
p²+5p = -6
p²+5p+(5/2)² = -6+(5/2)² --- By using the format of a²+2ab+b²=(a+b)², add (5/2)² on both sides
(p + 5/2)² = -6 + (5/2)²
(p + 5/2)² = -6 + 25/4
(p + 5/2)² = 1/4
p + 5/2 = +- sqrt(1/4)
p + 5/2 = +- (1/2)
p + 5/2 = 1/2 or p + 5/2 = -1/2
p = 1/2 - 5/2 or p = -1/2 - 5/2
p = -4/2 or p = -6/2
p = -2 or p = -3


2)2c²+c=5
c² + c/2 = 5/2 --- Multiple both sides by 1/2
c² + c/2 + [(1/2)/2]²= 5/2 + [(1/2)/2]² By using the format of a²+2ab+b²=(a+b)², add [(1/2)/2]² on both sides
c² + c/2 + (1/4)²= 5/2 + (1/4)²
(c + 1/4)² = 5/2 + 1/16
(c + 1/4)² = 41/16
(c + 1/4) = +- Sqrt(41/16)
c = Sqrt(41/16) - 1/4 or c = -Sqrt(41/16) - 1/4
c = 1.35 or c = -1.85 (both answers are corrected to 3 sig. fig.)

Answer 6

1. p^2 + 5p + 6 =0
or, p^2 +2p + 3p + 6 = 0
or, p(p+2) + 3(p+2) = 0
or, (p+2)(p+3) = 0
=> either p+2 =0 or p+3 = 0
=> either p= -2 or p = -3

2. 2c^2 +c = 5
here A = 2, B = 1 & C = -5
D = B^2 - 4AC
= 1^2 - 4(2)(-5)
2. 2c^2 +c = 5
here A = 2, B = 1 & C = -5
D = B^2 - 4AC
= 1^2 - 4(2)(-5)
= 1 + 40 = 41

NOW x = (-B + √D)/2A OR ( -B - √D)/2A

= (-1 + √41)/4 OR (-1 - √41)/4

Answer 7

1) p²+5p+6=0
Find a set of number where if you multiply will give 6, but added will give 5.
6 is 2 x 3 or 1 x 6 but only 2 x 3 when added will give 5 so now just factor them
(p + 2)(p + 3) = 0
the equation will be 0 if p + 2 = 0 or p + 3 = 0
hence, p = -2, -3


2)2c²+c=5
Rewrite
2c²+c - 5 = 0
then
based on ac²+bc + g = 0
roots are
[-b + sqrt(b^2 - 4ag)]/(2a), [-b - sqrt(b^2 - 4ag)]/(2a)
[-1 + sqrt(1 + 40)]/4 , [-1 - sqrt(1 + 40)]/4

c = -1/4 + [sqrt(41)]/4 , -1/4 - [sqrt(41)]/4

Answer 8

Because 0^0 is 1. Anything to the 0th power is 1

Answer 9

I have a feeling English frustrateS you too..