The Math Thread

[Anteroinen]

I had a math problem in my math test, which, alas, I couldn't solve on time. It went like this: with which values of the parametre a does the line y=x+a not cut the parabola y=-½x^2. I drew the problem out, and from that figured I could've calculated the lowest possible value of a, if I took the equation for the tangent of the parabola with x as the first term, but was unable to figure out how the Hell was I to do that. I suppose just making x+a > -½x^2 could've worked but it just got stuck with a<½x^2+x and then I run out of time.

So please enlighten me on how silly I was please.

[dVanDaHorns]

I had a math problem in my math test, which, alas, I couldn't solve on time. It went like this: with which values of the parametre a does the line y=x+a not cut the parabola y=-½x^2. I drew the problem out, and from that figured I could've calculated the lowest possible value of a, if I took the equation for the tangent of the parabola with x as the first term, but was unable to figure out how the Hell was I to do that. I suppose just making x+a > -½x^2 could've worked but it just got stuck with a<½x^2+x and then I run out of time.

So please enlighten me on how silly I was please.

Taking the derivative of y=-½x^2, you get y'=-x. The first point on the parabola where y=x+a can cross is thus when y=x+a is tangental to the parabola, or when the slope equals 1. Plugging that into the derivative, you get:
1=-x, x=-1
y=-½(-1)^2=-½, so the first point of intercept occurs when a is such that y=x+a passes through the point (-1,-1/2).
Plug those in, and solve for a.
-1/2=-1+a, a=1/2
Since the parabola curves downward, the line y=x+a does not cut the parabola for a>0.5

[Anteroinen]

I had a math problem in my math test, which, alas, I couldn't solve on time. It went like this: with which values of the parametre a does the line y=x+a not cut the parabola y=-½x^2. I drew the problem out, and from that figured I could've calculated the lowest possible value of a, if I took the equation for the tangent of the parabola with x as the first term, but was unable to figure out how the Hell was I to do that. I suppose just making x+a > -½x^2 could've worked but it just got stuck with a<½x^2+x and then I run out of time.

So please enlighten me on how silly I was please.

Taking the derivative of y=-½x^2, you get y'=-x. The first point on the parabola where y=x+a can cross is thus when y=x+a is tangental to the parabola, or when the slope equals 1. Plugging that into the derivative, you get:
1=-x, x=-1
y=-½(-1)^2=-½, so the first point of intercept occurs when a is such that y=x+a passes through the point (-1,-1/2).
Plug those in, and solve for a.
-1/2=-1+a, a=1/2
Since the parabola curves downward, the line y=x+a does not cut the parabola for a>0.5

Goddammit I was so close T_T. I even took the derivative once.

[dVanDaHorns]

...but it just got stuck with a<½x^2+x and then I run out of time.

Alternatively, using the method you were using, all you would have to do was solve for x, and then figure out what values of a would generate a complex x.
x^2+2x+2a=0
Using the simplified quadratic formula,
-1+/- sqrt(1-2a)
as long as (1-2a) is negative, the two will not intersect.
Also yielding a>0.5

Edit: don't worry, man! It happens to us all, even in the best of times!

[Anteroinen]

...but it just got stuck with a<½x^2+x and then I run out of time.

Alternatively, using the method you were using, all you would have to do was solve for x, and then figure out what values of a would generate a complex x.
x^2+2x+2a=0
Using the simplified quadratic formula,
-1+/- sqrt(1-2a)
as long as (1-2a) is negative, the two will not intersect.
Also yielding a>0.5

Edit: don't worry, man! It happens to us all, even in the best of times!

That was most likely what we were supposed to do, I remember working with the determinants early in the course.

[dVanDaHorns]

That was most likely what we were supposed to do, I remember working with the determinants early in the course.

Determinants? Nah, that's matrix methods, man. Which would also work. (But would be rather redundant, since this is a simple 2x2 system. )

[Anteroinen]

That was most likely what we were supposed to do, I remember working with the determinants early in the course.

Determinants? Nah, that's matrix methods, man. Which would also work. (But would be rather redundant, since this is a simple 2x2 system. )

Isn't the part under the square root in the quadratic formula called a determinant?

[dVanDaHorns]

That was most likely what we were supposed to do, I remember working with the determinants early in the course.

Determinants? Nah, that's matrix methods, man. Which would also work. (But would be rather redundant, since this is a simple 2x2 system. )

Isn't the part under the square root in the quadratic formula called a determinant?

...maybe? I am only aware of the definition of that term with respect to matrix methods, so I dunno.

[Isobel The Sorceress]

Isn't the part under the square root in the quadratic formula called a determinant?

It's called discriminant.

[Anteroinen]

Isn't the part under the square root in the quadratic formula called a determinant?

It's called discriminant.

Ah.