Solve for a When Parabola y=x^2-a(x+1)+3 Intersects x-Axis Once

[Asla]

Homework Statement

When the parabola y=x^2-a(x+1)+3 intersects the x-axis at one point then a=_ or =_.

Homework Equations

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

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.

 

[Saitama]

Homework Statement

When the parabola y=x^2-a(x+1)+3 intersects the x-axis at one point then a=_ or =_.

Homework Equations

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

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.

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

 

[LCKurtz]

y=x^2-ax-a+3
The equation already has two unknowns and hell breaks lose when I try to complete the square.

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

 

[Asla]

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

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.

 

[LCKurtz]

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.

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.

 

[Asla]

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.

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

 

[LCKurtz]

No, it isn't.

 

[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.

 

[verty]

3. Complete the square: y = ax^2 + bx + c

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

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 $4ac = b^2$.

 

[Asla]

No, it isn't.

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