Finding an X intercept

I just toally blanked out on my homework today, oh man. I need to find the X intercept, I just need to let Y equal 0.


y=x^2-4x+3

And this is what I have done

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

I know I should factor out the x^2-4x...

-3=x(x-4)

Could someone tell me if thats right, if its wrong, could u tell me the next step.

there's a formula to solve second degree polynomials:

ax^2+bx+c=0

then you have two solutions for x:
(-b + sqrt(b^2 - 4ac))/2a
and
(-b - sqrt(b^2 - 4ac))/2a

but only if b^2 - 4ac is not negative.

That one factorises as (x-1)(x-3) so the roots are 1 & 3. (Reasoning there is, that for the x intercept/s, y=0 so (x-1)(x-3)=0, ie either x-1 =0 so x=+1, or x-3=0 so x = +3)

With the simple case of a=1 like this, you choose factors that add to b and multiply to c. ie here, -1 + -3 = -4 and -1 x -3 = +3

Verify with the formula Ked gave:

sqrt(b^2 - 4ac) = sqrt (16-12) = sqrt 4 = 2

so the roots are (-(-4) +- 2)/2 = 6/2 = 3 or 2/2 = 1 as above.

Worth noting that if the portion in the square root is negative this eqn doesn't work as Ked said, and what it means is that there are no roots, ie the curve is sort of 'floating' clear of the x axis.

y=x^2-4x+3

y=(x^2-3x)+(-x+3)
y=x(x-3)-1(x-3)
y=(x-1)(x-3)
0=(x-1)(x-3)

x-1 = 0
x=1

or

x-3 = 0
x=3

learn this one early, amn, it is useful. two intercepts, as it is a parabola. good luck~

Quote:

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Originally posted by snakeeyes1000
two intercepts, as it is a parabola. good luck~

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Not always, though! If the square root part is zero (if my memory serves me, it's called the discriminant?) then the roots are the same, x₁ = x₂ = -b/2a. That's when the trough or peak just touches the x axis.

For the discriminant to be zero, b² = 4ac

discriminant = b[sup]2[/sup - 4ac

for the quadratic formula that's in a square root sign, so

delta
/_\ > 0 --> 2 solutions (intercepts)
/_\ = 0 --> 1 solution (intercept)
/_\ < 0 --> 0 solutions (intercepts)


sort of :p