Help solving quadratic equations?

Question

I need help with these quadratic equations...

3x^2 = 11x + 4

and


x^2 + 12x + 32 = 0


PLEASE HELP I AM LOST :)

Answer 1

3x^2 = 11x + 4
3x^2 - 11x - 4 = 0
(2 numbers to multiply to (3)(-4) = -12 and add to -11): -12, 1
3x^2 - 12x + x - 4 = 0
3x (x - 4) + (x - 4) = 0
(3x + 1)(x - 4) = 0

3x+1 = 0
x = -1/3

x - 4 = 0
x = 4

2.
x^2 + 12x + 32 = 0
(2 numbers that multiply to 32 and add to 12) : 8, 4
(x + 8)(x + 4) = 0

x = -8
x = -4

Answer 2

Rewrite the first one:
3x^2 - 11x - 4 = 0
factor it: (3x+1)(x-4)=0 so that x=-1/3 or +4
You can also use the quadratic formula, but factoring, when it is fairly easy is easier with fewer steps...if you can see the factors easily...on a test, don't waste time on it if you don't see it immediately and use the quad

x^2 + 12x +32 = 0
(x+8)(x+4) = 0 and x=18,-4
This one was really easy

I hope that I have helped you

Answer 3

OK so lets go back to the quadratic equation formula:

if your quadratic equation is in the form

ax^2 + bx +c = 0

Then you can use the fomula

x = (-b + Squareroot(b^2 - 4ac)) / 2a
x = (-b - Squareroot(b^2 - 4ac)) / 2a
(see a cleaner version below)
http://en.wikipedia.org/wiki/Quadratic_equation

Find out what a b and c are and plug them into the quadratic equation formula.

In the first problem, try rewriting that thing to look like
ax^2 + bx + c = 0
by rearranging the terms.

For the second one, its already in that form so just plug it in. There is no 'a'! you may say. Its just x^2!!

ah but remember that x^2 is the same as 1x^2, so a = 1 in this case.

Answer 4

Make one side of the equation zero.
Then factor the other side.

3x^2 = 11x + 4
3x^2 - 11x - 4 = 0
(3x + 1)(x - 4) = 0
3x = 1
x = 1/3
x = 4


Question 2: Just factor
(x + 4)(x + 8) = 0
x = -4
x = -8

Answer 5

3x^2=11x+4
3x^2-11x-4=0

substitute numbers into the quadratic equation:
a=3
b=-11
c=-4

x=4 or x= -1/3


x^2+12x+32=0

substitute numbers into the quadratic equation:
a=1
b=12
c=32

Answer: x= -4 or x=-8

Email me if you need the work shown...

Answer 6

3x^2 = 11x + 4 subtract 11x+4 from each side
3x^3-11x-4=0
(3x+1)(x-4)=0
3x+1=0
3x=-1
x=-1/3
x-4=0
x=4
roots are x=4, -1/3

x^2 + 12x + 32 = 0
(x+8)(x+4)=0
x+8=0
x=-8
x+4=0
x=-4
roots are x=-4, -8

Answer 7

set up the first equation to be in standard form

3x^2-11x-4=0

a=3
b=-11
c=-4

remember the formula? -b+ or - rad b^2 - 4ac all over 2a

substitute:

-(-11) +or - rad (-11)^2 -4(3)(-4) all over 2(3)

can you finish it?

The other is already in standard form, and easier since you have no double negs to bother with.... Helpful?

If I didn't make any screwy mistakes, the ans to the first one is

11+ or - rad 121+48 all over 6
11 + or - rad 169 all over 6
11 + sq rt of 169 all over 6
11 + 13 all over 6
24 over 6 or
4
and finish with subtracting rad (sq. rt of 169) 13 all over 6
or -1/3 so it has two answers, since, as you know, x is squared