# Is this question possible without having learnt derivatives?

1. Dec 31, 2009

### ottoic

nvm got it

Last edited: Dec 31, 2009
2. Dec 31, 2009

### rock.freak667

Right, well when you reach here, you will get

$$3x^2+6x-(1+b)=0$$

Now if we solve for x using the quadratic equation we'll get

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

If the line intersects the curve, you'll get two real values for x right? (in the form a±b)

Now for a tangent, the line intersects the curve once. So what condition should you put so that in the quadratic formula above, to get only value for x?

3. Dec 31, 2009

### rphenry

Suggestion (without working it all out): Look for a value of b such that the line intersects the parabola at one point.

From where you were:

0=3x^2+6x-1-b

Isolate b:

b = 3x^2+6x-1

Solve the quadratic equation for b. Does that work?

4. Dec 31, 2009

### HallsofIvy

No, tha'ts not a quadratic equation "for b"- and it is already solved for b.

rockfreak667 suggested the right method: the equation $ax^2+ bx+ c= 0$ has a single solution if and only if the "discriminant", [itex]b^2- 4ac= 0[/math]. For this equation that would be 36+ 12(b+1)= 0. Solve that equation for b.