Can you please help me with quadratic equations?

Question

I have these two problems and I need help solving them. Can you please show me your steps to help me solve these problems.

x^2 - 5x +6 = 0

and

x^2 -x =12

I know the answer already I jsut need the steps to solve this form of quadratic equation

Answer 1

(x-3)(x-2)=0
x-3 =0 or x-2 = 0
x = 3 or x = 2

x^2-x-12=0
(x-4)(x+3)=0
x-4=0 or x+3=0
x=4 or x=-3

Answer 2

The quadratic formula for the first one will give you a =1, b= -5, c=6.
Applying the formula:
x =(5 +- sqrt[5² - 4(1)(6)] )/2(1)
x = (5 +- sqrt(25-24))/2
x =( 5 +- sqrt(1))/2
x =(5+- 1)/2

so the two solutions are:
(5 + 1)/2 and (5 - 1)/2
or, 3 and 2.

For the second one, bring the 12 to the other side, to make it look familiar.
x^2 - x - 12 = 0. You can see that a=1, b= -1, and c = -12.
applying the quadratic formula:

x = [ 1 +- sqrt(1^2 - 4(1)(-12)) ]/2(1)
x = [1 +- sqrt(1 + 48)]/2(1)
x = [1 +- sqrt(49)]/2
x = [1 +- 7]/2

so the two solutions are
[1 + 7]/2 and [1 - 7]/2
or 4 and -3.

Answer 3

x^2 - 5x +6 = 0

This factors easily

(x - 2)(x - 3) = 0

The statement is true when either x - 2 = 0 or x - 3 = 0

x = 2, 3

x^2 - x = 12
x^2 - x - 12 = 0

This also factors easily

(x - 4)(x + 3) = 0

x = 4, -3

Yu must have been doing some work factoring trinomials. For a good factoring lesson, see the info at the link below

Answer 4

ok, gonna be hard to explain it

of picture it like this

1) a - b + c

so, what multiplys to ur c, and adds to ur b (multiplies to 6, and adds to -5)

-2 times -3 = -6, and -2 + -3 = 5
so then u do
x-2= 0 or x-3 = 0
so x = 2 or x = 3
then u just plug it in to see which = 0




for the next one, do the same exact thing, u just hav to get the 12 on the same side as the other ones. if u cant follow the a b c rule, then use the quadratic formula

if u need more help, u can email me

Answer 5

You have to use the quadratic formula or factor
1.)
x^2-5x+6=0

(x-3)(x-2)=0
x=3 and 2 (3,0) and (2,0)

2.)
a=1 b=-1 c=0
(-1)^2=1 4(1)(0)=0 1-0=1
square root of 1 is 1
-(-1)plus/munus 1 1plus/minus1= 2 and 0 1+1=2 1-1=0
2a = 2(1) = 2
2/2 = 1 and 0/2=0
Solution
x=1 (1,0)
x=0 (0,0)

Answer 6

1. x^2-5x+6=o
2. (x^2-5x)+6=0
3. (x^2-5x +(5/2)^2-(5/2)^2)+6=0
4. (x-5/2)^2+6-(5/2)^2=0
5. (x-5/2)^2+6-(25/4)=0
6. (x-5/2)^2+24/4-25/4=0
7. (x-5/2)^2 -1/4=0
8. 1/4= (x-5/2)^2=0
9. sq. root of 1/4=x-5/2
10. x=sq. root of 1/4 +-5/2
11. x=sq root of 1(divide 2) + - 5/2
12. sq root of 1 + - 5 all divided by 2

Answer 7

you use the quadratic formula

x=(-b +- square root of b^2-4ac)/2a
a=1, b=-5, c=6
so,
x=(-(-5) +- square root of 25-24)/2*1
x=(5 +- square root of 1)/2
x=(5+1)/2 and x=(5-1)/2
x=6/2 and x=4/2
x=3 and x=2


2nd equation:
x=(-(-1)+- square root of 49)/2*1
x=(1+-7)/2
x=8/2 and x=-6/2
x=4 and x=-3

there you go!

Answer 8

x^2 - 5x + 6 = (x - 3)(x - 2)
x = 3 or 2

--------------------

x^2 - x = 12
x^2 - x - 12 = 0
(x - 4)(x + 3) = 0
x = 4 or -3

Answer 9

for each of these just plug the a, b, and c values into the quadratic formula and solve for x.

for the first one:
a = 1
b = -5
c = 6

for the second one:
a = 1
b = -1
c = -12

and then simplify and there are your answers for x.

Answer 10

1)

X^2 - 5x + 6 =0
x^2 -3x -2x + 6 = 0
x(x-3) -2(x-3)=0
(x-2)(x-3)=0
x=2,3

2)

x^2 - x -12=0
x^2 -4x +3x -12 = 0
X(x-4) +3(x-4)=0
(x-4)(x+3)=0
x=4,-3