help with this math problem?

Question

okay i cant seem to understand how to solve this

y=x^2-4x+3 when y=2x-2

i believe its somethin like this x^2-6x+5, am i right?? but then how i solve it, quadratic formula? can some1 show me,

listing steps is appreciated. thanks

Answer 1

2x - 2 = x^2 - 4x + 3
Set one side equal to 0
0 = x^2 - 6x + 5
Factor
0 = (x - 5)(x - 1)
Set each factor equal to 0
0 = x - 5 0 = x - 1
Solve
5 = x 1 = x

Answer 2

Substitute.

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

2x = x^2 - 4x + 5

x^2 - 6x + 5 = 0 So basically, you are right.

Solving it from here has two possibilities:

Factor and set each to 0.

x^2 - 6x + 5 = (x-5)(x-1) --> x-5=0 and x-1=0. Then x=1, x=5.

Use the quadratic formula.

6 +/- sqrt(36-4*1*5)
________________

2

Then the answer is x = 3 +/- 2, or 1 and 5.

Answer 3

You're correct, the equation becomes x^2-6x+5 = 0, which you obtain by moving y to the other side and collecting like terms. Then you can either use the quadratic formula to find the roots, or factor. To factor, you try to find two numbers a and b so that a*b = 5 and a+b= 6. Once you find these numbers, the equation factors as (x-a)*(x-b) = 0, and so a and b are the roots.

Answer 4

You are nearly there. Consider that when you substituted for y and then collected like terms you really ended up with :-

x^2 - 6x + 5 = 0.

The solution of this quadratic is simple, ask 'what two factors of '+5' have a sum of '-6' ?. Simply put, if these two numbers are 'a' and 'b', ab = +5, and a + b = -6. Look at the equation to see their origins. But further, the two quadratic factors must have a product of zero when x is made to equal either '-a' or '-b'.

Clear? Probably not.

If the quadratic factors are Q1 and Q2,

(Q1)(Q2)=0, for this to be true, either Q1 or Q2 must be zero (remembering that 'x' can have only one value at any time) and so....

(x-5)(x-1)=0 is the quadratic equation, and for this to be true,
x = 5 or 1.

Now, go away and look this up in your texts and speak with your teacher!

Answer 5

So you want to solve for x and you have these equations:

y= x^2 - 4x + 3
y= 2x - 2

Set them equal to each other and move everything to one side:

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

x^2 - 6x + 5 = 0

Now we can factor this:

(x - 5)*(x - 1) = 0

Now we have two equations for x:

x - 5 = 0
x - 1 = 0

Solving these two gives us:

x = 5
x = 1

So x can equal 5 or 1

Answer 6

Hmm....a simultaneous equation?

y = x^2 - 4x + 3
y = 2x -2

x^2 - 4x +3 = 2x -2
x^2 - 4x -2x + 3 + 2 = 0
x^2 - 6x + 5 = 0
(x - 5)(x - 1) = 0
x = 5 or x = 1

Answer 7

y = x^2 - 4x + 3
y = 2x - 2
So,
2x - 2 = x^2-4x+3
x^2-4x+3 = 2x-2
x^2 - 4x + 3 + 2 = 2x
x^2 - 4x + 5 = 2x
x^2 - 4x - 2x + 5 = 0
x^2 - 6x + 5 = 0
x^2 -x - 5x + 5 = 0
x(x - 1) -5(x - 1) = 0
(x - 5)(x - 1) = 0
Either x - 5 or x - 1 = 0
Put x - 5 = 0
x = 5
Put x - 1 =0
x = 1
x = 5,1 are the two possible values of 'x' that satisfy the equation

Answer 8

x=the square root of 5/6, you take the 5 on the other side which is how you get -5, then divide the 6 from both sides, which is how you get -5/6, then take the add a negative to both sides you will have -x^2, then square root both sides to x alone on the one side, therefore you have, square root of 5/6. it should be .9128709 or -.9128709. The other option is to try Math.com and look at the algebra equations, great site.

Answer 9

1)-----y=x^2-4x+3
2)-----y=2x-2
put 2 into 1

2x-2=x^2-4x+3
x^2-6x+5=0
(x-1)(x -5)=0
x=1 or x = 5

Answer 10

http://www.purplemath.com/modules/factquad.htm
use this website to understand how to find the factors

(x-5)(x-1)
x=5 or x=1