Homework Help: Y = x^2-a(x+1)+3

1. Jun 20, 2013

Asla

1. The problem statement, all variables and given/known data
When the parabola y=x^2-a(x+1)+3 intersects the x-axis at one point then a=_ or =_.

2. Relevant equations
The equation of parabola y^2=4px,x^2=4py depending on the orientation

3. The attempt at a solution
Since the parabola meets the x axis once, I assume that to be the vertex.I try to rearrange the function into an appropriate form but I get stuck here.
y=x^2-ax-a+3
The equation already has two unknowns and hell breaks lose when I try to complete the square.

2. Jun 20, 2013

Saitama

What is the condition for a quadratic to have a single root?

3. Jun 20, 2013

LCKurtz

How exactly does all hell break loose? What do you get?

4. Jun 20, 2013

Asla

I get something to this effect;
y+1/2a-3=(x-1/2a)^2
I get stuck because the y part cannot be factorized but I am again thinking that the I could equate 1/2a-3=0 because apparently the left side should not have a constant since the vertex is along the x-axis.That could give me one of the values of a as 6.

5. Jun 20, 2013

LCKurtz

When you write 1/2a do you mean a/2 or 1/(2a). You should use parentheses. In any case, show your steps on completing the square because what you have isn't correct.

6. Jun 20, 2013

Asla

Oh my mistake,
I got this y+(1/4)a^2-3=(x-(1/2)a)^2
Is that what you have?

7. Jun 20, 2013

LCKurtz

No, it isn't.

8. Jun 20, 2013

verty

You may want to revisit parabolas and completing the square, you are missing some knowledge there.

Some practice:
1. Complete the square: 2x^2 + 5x - 4 = 0
2. Complete the square: y = 2x^2 + 5x - 4
3. Complete the square: y = ax^2 + bx + c

Your answer for #3 should be $y = a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}$.
Then...

No, what I said after this was not correct. But this is good practice still.

Last edited: Jun 20, 2013
9. Jun 20, 2013

verty

If you let y = 0 and solve for x, you get the well-known formula for the roots of a parabola. But already in this form it is clear that something strange happens when $4ac = b^2$.

10. Jun 20, 2013

Asla

I redid it.
I got this:
y+(1/4)a^2+a-3=(x-(1/2)a)^2
and solved a to be -6 or 2