Homework Help: 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.