How do you solve 8p ^ { 2} + 38p + 62= - 8p + 6?

2 Answers
Oct 22, 2017

p = -1.75 or p = -4

Explanation:

1) Rearrange so that one side = 0:

8p^2 + 38p + 62 + 8p - 6 = -8p + 6 + 8p - 6
(Adding 8p - 6 to both sides)

8p^2 + 46p + 56 = 0

2) Use the quadratic formula:

p = (-b +- (sqrt(b^2 - 4ac)))/(2a)

Where a = 8, b = 46, c = 56.

p = (-46 +-(sqrt(46^2 - 4 * 8 * 56)))/(2*8)

p = (-46 +-sqrt(324))/16

p = -1.75 or p = -4

Oct 22, 2017

p=-4 or p=(-7)/4

Explanation:

8p^2+38p+62=-8p+6

Subtract (-8p+6) from both sides.

8p^2+38p+62-color(red)((-8p+6))=-8p+6-color(red)((-8p+6))

8p^2+46p+56=0

Divide both sides by 2

(8p^2)/color(red)2+(46p)/color(red)2+58/color(red)2=0/color(red)2

4p^2+23p+28=0

Now we have to find two numbers whose product is equal to 28xx4 (28xx4=112) and add up to 23.

The two numbers are 16 and 7

So 4p^2+23p+28=0 can be written as 4p^2+16p+7p+28=0

(4p^2+16p)+(7p+28)=0

4p(p+4)+7(p+4)=0

(p+4)(4p+7)=0

So (p+4)=0 and (4p+7)=0

color(red)1.p+4=0

p=-4

and

color(red)2.4p+7=0

4p=-7

p=(-7)/4