
Moar Quadratics!
1. Find an equation for the straight lines which pass through the point (1, 2) and are tangent to the parabola with equation y = x^{2}.
Would my two equations be y = x^{2} and 2 = m + c? Where do I go from here? Usually I would make equation 1 equal equation 2, but y is not equal to 2.
2. Solve the following for x by completing the square:
2x^{2}  4x + 1 = 0
What I tried was:
2(x^{2 } (4/2)x + (1/2)) =0
(x  1)^{2} + 1 + (1/2)  1 = 0
(x  1)^{2 } (1/2) = 0
x  1 = sqrt(1/2) = 0
x = 1 + sqrt(1/2)
Thanks in advance.

Re: Moar Quadratics!
Let's work the first problem before moving to the second. Using the pointslope formula, with unknown slope $\displaystyle m$, what is the family of lines which pass through the given point?

Re: Moar Quadratics!
y  y_{1} = m(x  x_{1})
y + 2 = m(x  1)
y = mx  m + 2
y = m(x  1) + 2

Re: Moar Quadratics!
Yes, this is correct:
$\displaystyle y(2)=m(x1)$
and so:
$\displaystyle y+2=m(x1)$
This is what you correctly found,
Now subtract 2 from both sides to solve for y...I think you just got a little careless when you did this. :D

Re: Moar Quadratics!
:D Yep, I see it.
y = m(x  1)  2
Would we then do something like...?
x^{2} = m(x  1)  2
x^{2} + m(x  1)  2 = 0
If we were to solve this by equating the discriminant to 0, would the b constant of said discriminant be?

Re: Moar Quadratics!
First, you want to write the quadratic in standard form $\displaystyle ax^2+bx+c=0$. I recommend distributing the m, and then multiplying through by 1. What do you get?

Re: Moar Quadratics!

Re: Moar Quadratics!
Yes, so equate the discriminant to zero and solve for m...what do you find?

Re: Moar Quadratics!
What is the b value of the discriminant? Is it m?

Re: Moar Quadratics!
It is m actually, but this won't affect the computation of the discriminant, but it is good to recognize that it we have:
$\displaystyle (1)x^2+(m)x+(m+2)=0$
This way we can easily see that:
$\displaystyle a=1,\,b=m,\,c=m+2$.

Re: Moar Quadratics!
Ah, I see. I'll complete this equation later and post my results.

Re: Moar Quadratics!
m^{2}  4m  8 = 0
Therefore m = 2 + 2sqrt3.
So do we then just substitute this m value back into y = mx  m  2?

Re: Moar Quadratics!

Re: Moar Quadratics!
Awesome.
Where did I go wrong with the following equation?
2(x2  (4/2)x + (1/2)) =0
(x  1)2 + 1 + (1/2)  1 = 0
(x  1)2  (1/2) = 0
x  1 = sqrt(1/2) = 0
x = 1 + sqrt(1/2)

Re: Moar Quadratics!
When you took the square root of both sides, you did not consider the negative root.
If $\displaystyle x^2=y$ then $\displaystyle x=\pm\sqrt{y}$