Quadratic Definition and 965 Threads

My solution: 1. ##K(a^2) \subseteq K(a)##. 2. ##a## is zero of the quadratic polynomial ##X^2 - (a^2)##, i.e., ##[K(a) : K(a^2)] \leq 2##. 3. It is not 2 because ##[K(a) : K] = [K(a) : K(a^2)][K(a^2) : K]## is odd. 4. Thus, it is 1, and hence ##K(a) = K(a^2)##. Is this solid enough?

Given the following equation: R = ((Q-P) / |Q-P|) ⋅ V where Q, P, and V are 3 dimensional vectors, R is a scalar, "⋅" denotes the dot product, and |Q-P| is the magnitude of Q-P. Assuming Q, V, and R are known and given 3 independent equations with different values for Q, V, and R that...

For example, let's say the quadratic curve ax^2 + bx + c intersects the x-axis at x=-5, x = 3 Why is it we can say the equation of the curve is then (a)(x+5)(x-3) ? how does this manipulation come about, in particular, the (a) coefficient. Why can we just slot (a) into the factors (x+5)(x-3) ...

After the news of Peter Higgs's death, I was thinking of the Higgs field and boson, how central they are to physics now, and the remaining mysteries associated with them. One such is the meaning of the mass actually observed for the Higgs boson. In the mainstream of theoretical physics, the...

##y+3=3\sqrt{y+7}## Square both sides: ##\Rightarrow y^2+6y+9=9y+63## ##\Rightarrow y^2-3y-54=0## ##\Rightarrow (y-9)(y+6)=0## ##y=9, -6## But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

Using the complete the square method I got; -3+root8 or -3-root8 But using the quadratic formula (to check my answer) I got; -3+root 10 or -3-root 10 I've checked both answers several times but can't get to the bottom of it :)

I was trying to show that ##sin(x-y) = sin(x)cos(y)-cos(x)sin(y)## using Pythagoras' theorem and ##cos(x-y)=cos(x)cos(y)+sin(x)sin(y)##. I have: $$sin^2(x-y)=1-cos^2(x-y)$$ $$sin^2(x-y)=1-(cos(x)cos(y)+sin(x)sin(y))^2$$...

I am struggling to rederive equations (61) and (62) from the following paper, namely I just want to understand how they evaluated terms like ##\alpha\epsilon\alpha^{T}## using (58). It seems like they don't explicitly solve for ##\alpha## right?

Q. 21 Allow me to post using phone... Will amend/ show working when i get hold of computer.In my working I have ##5x^2 -cx^2-4x-2<0## ##56-8c<0## ... ##c>7## Textbook says ##7>c## ...who's fooling who here? 😊 I have seen my mistake. I am wrong...put in wrong inequality sign... Should be...

Now i just need some clarification; we know that quadratic equations are equations of the form ##y=ax^2+bx+c## with ##a,b## and ##c## being constants and ##x## and ##y## variables. Now my question is... can we also view/look at ##x=y^2+2y+1## as quadratic equations having switched the...

The Solution of the Quadratic Equation By Differentiation Method

Looking at the proof of the Schwarz inequality in Margenau and Murphy, you will see what I attached. Gamma is asserted to be positive (OK). Given that the usual "quadratic form" solution would read "-(B+B*) .....". The sign does not seem correct to me as shown. In a fact B+B* = 2Re(B) and...

Im a having trouble understanding how this exactly works. $$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases. 1. ## x^2-4 >= 0 ## and 2. ## x^2-4...

I noticed that ##V(\phi)## has nonzero minima, therefore I found the stationary points as ##{{\partial{V}}\over{\partial\phi}}=0##, and found the solutions: $$\phi^0_{1,2}=-{{m}\over{\sqrt{\lambda}}}\quad \phi^0_3={{2m}\over{\sqrt{\lambda}}}$$ of these, only ##\phi^0_3## is a stable minimum...

I was given a problem to solve that goes like this ##\frac{3}{|x+3|-1}\geq |x+2|## . I got the correct solution for all possible cases and here they are; for ##|x+3|\geq0## and ##|x+2|\geq## i got ##x\epsilon <-2, -2\sqrt{3} ]## and for ##|x+3|\leq0## , ##|x+2|\leq0## I got ##x\epsilon [-5...

https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/ An interesting article about solving ax2 + bx + c = 0 = (x-R)(x-S), where R and S are the roots. ## x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} ## In my classes, we were never 'spoon fed' any formula, but...

Can someone please tell me where I am wrong. I am learning how to write ##a^{2} + bx + c## in this form ##f(x)= a(X-X_0)^{2} +Y_0##. The method used in my textbook is a reduction to the perfect square. And it goes like this: ##f(x)=ax^2+bx+c## ##=a[x^2+\frac{b}{a}x]+c## ##=a\left [...

I am a bit confused, so if anyone can explain to me which way is right I would be very thankful. I think that the way in pic 1 is right because of the properties written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2 where he didn't...

I learned that for a bilinear form/square form the following theorem holds: matrices ## A , B ## are congruent if and only if ## A,B ## represent the same bilinear/quadratic form. Now, suppose I have the following quadratic form ## q(x,y) = x^2 + 3xy + y^2 ##. Then, the matrix representing...

Substitute (a,b,c) into z=y^2-x^2: c=b^2-a^2 Substitute the parametric equations of L1 into the equation of the hyperbolic paraboloid in order to find points of intersection: z=y^2-x^2 c+2(b-a)t=(b+t)^2-(a+t)^2=b^2-a^2+(b-a)t c=b^2-a^2

I am refreshing on this...Have to read broadly...i will start with (b) then i may be interested in alternative approach or any correction that may arise from my working. Cheers. Kindly note that i do not have the solutions to the following questions... For (b), we know that, say, if ##x=α##...

Ever made a simple model that fits a quadratic function? Tweaking the a, b and c constants to fit new observed data is a bit of a pain. When I was a grad. student I came up with the following simple quadratic rearrangement that uses the intercept (Yo) and the values of x and y that define the...

Let the roots of the given quadratic equation be ##x=α## and ##x=β## then our quadratic equation will be of the form; $$x^2-(α+β)x+αβ$$ It follows that ##(α+β)=(r+is)## and ##αβ=4##. We are informed that ##α^2+β^2=6i ## then $$6i=(r+is)(r+is)-8$$ ... $$8+6i=(r^2-s^2)+2rsi$$ solving the...

For part (i), ##(x-α)(x-β)=x^2-(α+β)x+αβ## ##α+β = p## and ##αβ=-c## therefore,##α^3+β^3=(α+β)^3-3αβ(α+β)## =##p^3+3cp## =##p(p^2+3c)## For part (ii), We know that; ##tan^{-1} x+tan^{-1} y##=##tan^{-1}\left[\dfrac...

For part a, We have ##α+β=b## and ##αβ =c##. It follows that, ##(α^2 + 1)(β^2+1)=α^2β^2+α^2+β^2+1)## =##α^2β^2+(α+β)^2-2αβ +1## =##c^2+b^2-2c+1## =##c^2-2c+1+b^2##...

ok this was posted on LinkedIn and sure it has already be answered but usually these types of problems are resolved by way too many steps so just wanted to proceed with this without looking at previous attempts my first reaction was to get a CD but would introduce a bigger problem however...

Find the question below; Find my working below; I hope i understood what the question was asking...you may confirm. Cheers guys

My guess would be to do an integral of the form $$\frac{\int d^4k}{(2\pi)^4}k^2(\frac{1}{(k^2-m^2+i\epsilon)}-\frac{1}{k^2-\Lambda_1^2+i\epsilon})(\frac{1}{(k^2-m^2+i\epsilon)}-\frac{1}{k^2-\Lambda_2^2+i\epsilon})$$ before Wick otating and integrating. Any help is appreciated. Thanks.

see the textbook problem below; see my working to solution below; i generally examine the neighbourhood of the critical values in trying to determine the correct inequality. My question is "is there a different approach other than checking the neighbourhood of the critical values"? In other...

So the top of the structure is a triangle with height x. and the height of the rectangle is 2 + x, and the length is 3x. I'm unsure where to go from here. I tried using the formula and getting 3x^2 + 6x +x = 3x^2 + 7x -336 =0 I applied the quadratic formula but it gave me non-integer solutions...

I subtract 5 from both sides to get 7x^2 = -5 Then I divide both sides by 7 to get -5/7. I then take the square root to get x = sqrt of the imaginary unit i 5/7 then ##\pm { i \sqrt \frac 5 7}## The quadratic formula on the other hand gets me a different answer, the discriminant = -140 which...

I am confused in (iia) and (iib). If $x^4 +( \alpha - 1) x^2 + \alpha + 2 = 0$ has real roots that means $y^2 + ( \alpha -1) + \alpha + 2 =0 $ should have at least one non-negative root. This means product of roots of (2) can be greater or less than zero...But I'm not able to comment on sum of...

https://getpocket.com/explore/item/mathematician-finds-easier-way-to-solve-quadratic-equations This seems to just be the quadratic formula in a transposed way. :rolleyes:

Does this curve look quadratic or exponential

Hello: I'm not sure if there's an accepted canonical form for a quadratic equation in two (or more) variables: $$ax^2+by^2+cxy+dx+ey+f=0$$ Is it the following form? (using the orthogonal matrix Q that diagonalizes the quadratic part): $$ w^TDw+[d \ \ e]w+f=0$$ $$w^TDw+Lw+f=0$$ where $$...

Given equation and conditions: ##\boldsymbol{x^2+2(k-3)x+9=0}##, with roots ##\boldsymbol{(x_1,x_2)}##. These roots satisfy the condition ##\boldsymbol{-6<x_1,x_2<1}##. Question : ##\text{What are the allowable values for}\; \boldsymbol{k}?## (0) Let me take care of the determinant first...

The equation $$(a-1)x^2-4ax+4a+7=0$$ with a is a whole number has positive roots. If $$x_1>x_2$$ then $$x_2-x_1=...$$ A. –8 B. –5 C. –2 D. 2 E. 8 Since the equation has positive roots then $$x_1>0$$ and $$x_2>0$$ thus $$x_1+x_2>0$$ and $$x_1x_2>0$$ $$x_1+x_2>0$$ $$\frac{-(-4a)}{a-1}>0$$...

On simplifying the given equation we get, x^2-x-1=0 and using the quadratic formula we get x=(1+√5)/2 and x=(1-√5)/2 Now, as the formula suggests, there are two possible values for x which satisfies the given equation. But now, if we follow a process in any general calculator by entering...

A golfer hits a tee shot into the rough and the ball stops approximately 120 yds from the green. There is a tree located 40 yds from the ball, directly in the path of the shot. The golfer decides to try to hit the ball over the tree. The path of the shot can be modeled by the equation h =...

I can replace f(x) with x - x^2. Should I plug (x/2) into f(x)? How about (x - 1/2) into f(x)? I need the set up.

Let me start by pasting the question as it appears in the text :My Attempt : Given equation : ##\boldsymbol{2x^2+mx+m^2-5 = 0}##. For the roots of this equation to be real, the discriminant : ##m^2-8(m^2-5) \ge 0\Rightarrow 7m^2-40\le 0\Rightarrow -\sqrt{\frac{40}{7}} \le m \le...

Given : The equation ##2x^2-(a^3+8a-1)x+a^2-4a = 0## with roots of opposite signs. Required : What is the value of ##a## ? Attempt : The roots of the equation must be of the form ##\alpha, -\alpha##. The sum of the roots ##0 = a^3+8a-1##. I do not know how to solve this equation. However...

Given : Equation ##x^2+(2m+1)x+(2n+1) = 0## where ##m \in \mathbb{Z}, n \in \mathbb{Z}##, i.e. both ##m,n## are integers. To prove : If ##\alpha,\beta## be its two roots, then they are not rational numbers. Attempt : The discriminant of the equation ##\mathscr{D} = (2m+1)^2 - 4(2n+1) =...

I have come across this strange method of solving degree 2 polynomials but I do not find the explanation provided to be very helpful. Here is the method description: "In the 16th century, mathematician Francois Viete solved quadratic equations by a unique substitution method. To solve an...

Given : The quadratic equation ##x^2+px+q = 0## with coefficients ##p,q \in \mathbb{Z}##, that is positive or negative integers. Also the roots of the equation ##\alpha, \beta \in \mathbb{Q}##, that is they are rational numbers. To prove that ##\boxed{\alpha,\beta \in \mathbb{Z}}##, i.e. the...

It is given that ##x_1, x_2\; \text{and}\; x_3## are roots of the equation ##ax^2+bx+c=0##, which are pairwise distinct. If indeed they are roots, we should have ##ax_1^2+bx_1+c= 0 = ax_2^2+bx_2+c= 0 = ax_3^2+bx_3+c= 0##. On subtracting the first two, we obtain ##a(x_1^2-x_2^2)+b(x_1-x_2) =...

Hi , I had to solve a quadratic equation , i got two roots as an answer ( ans= x1 / x2) , and now i need to use one of those answers to complete further tasks like finding y from x+y=c so i need to use x1 and x2 from roots , i was wondering if that's possible and how