<span>What is the yield to maturity on a simple loan for $1 million than requires a repayment of $2 million in five year time?
</span>
Answer:
See Explanation
Step-by-step explanation:
Given
![3x^2 = 6x](https://tex.z-dn.net/?f=3x%5E2%20%3D%206x)
Required
Why dividing by
is not reliable
The reason is that, the given expression is a quadratic equation and as such, it will most likely have two solutions.
So, when the expression is divided by
, the solution will be
, leaving out the other possible solution.
The right way is:
![3x^2 = 6x](https://tex.z-dn.net/?f=3x%5E2%20%3D%206x)
Subtract 6x from both sides
![3x^2 - 6x = 6x - 6x](https://tex.z-dn.net/?f=3x%5E2%20-%206x%20%3D%206x%20-%206x)
![3x^2 - 6x = 0](https://tex.z-dn.net/?f=3x%5E2%20-%206x%20%3D%200)
Factorize
![3x(x - 2) = 0](https://tex.z-dn.net/?f=3x%28x%20-%202%29%20%3D%200)
Split
![3x = 0\ or\ x - 2 = 0](https://tex.z-dn.net/?f=3x%20%3D%200%5C%20or%5C%20x%20-%202%20%3D%200)
Solve for x in both cases:
![x = 0\ or\ x = 2](https://tex.z-dn.net/?f=x%20%3D%200%5C%20or%5C%20x%20%3D%202)
from the provided focus point and directrix, we can see that the focus point is above the directrix, meaning is a vertical parabola and is opening upwards, thus the squared variable will be the "x".
keeping in mind the vertex is half-way between these two fellows, Check the picture below.
![\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \begin{cases} h = 7\\ k = 7\\ p = 2 \end{cases}\implies 4(2)(y-7)=(x-7)^2\implies 8(y-7)=(x-7)^2 \\\\\\ y-7=\cfrac{1}{8}(x-7)^2\implies \boxed{y=\cfrac{1}{8}(x-7)^2+7}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%20%3D%207%5C%5C%20k%20%3D%207%5C%5C%20p%20%3D%202%20%5Cend%7Bcases%7D%5Cimplies%204%282%29%28y-7%29%3D%28x-7%29%5E2%5Cimplies%208%28y-7%29%3D%28x-7%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-7%3D%5Ccfrac%7B1%7D%7B8%7D%28x-7%29%5E2%5Cimplies%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B8%7D%28x-7%29%5E2%2B7%7D)
Y=x^2-4x+8 and 4x+y=12 so y=12-4x, insert y in the quadratic to get
12-4x=x^2-4x+8
12=x^2+8
0=x^2-4=(x-2)(x+2)
So the solutions are x=2 and x=-2.
X = 3 and x = 4 will make the polynomial = 0
answer
C. 3 and 4