Question

I have taken a screening test today on a summer course for Linear Algebra. I would like to have the solution for question 4 and some hints on question 10.

Furthermore, I solved question 8 with angle sum identity but would like to know a way that wouldn't require it.

Answer 1

Here is the question booklet.

Answer 2

alpha and beta are the roots ,right?

Answer 3

Yep. I am thinking something like \[(-x+\alpha)(x-\beta)\] but couldn't finish it.

Answer 4

i dont think it is solved that way
now my memory is bit rusty,i solved these like 8 years ago?
i will google and fill in my blank spaces

Answer 5

the y intercept is 0,4 then one of the x's value is 4

Answer 6

when f(x) = 0

Answer 7

c is 4 but I couldn't figure the rest out.

Answer 8

how is c 4 O.o

Answer 9

\[
\begin{align*}
F(x)&=-x^2+bx+c\\
4&=-0^2+0b+c\\
c&=4
\end{align*}
\]

Answer 10

hey,no... one of x's values is 4,not whole function's value

Answer 11

Isn't the y-intercept of F(X) (0,4)?

Answer 12

yes

Answer 13

Doesn't F(0)=4 implies c=4? (or am I mad today?)

Answer 14

*shrugs*
okay am confused with these basics only
wait
@ganeshie8 how come he took f(0) = 4?

Answer 15

Isn't (0,4) a point on F(x)?

Answer 16

yes so c = 4

Answer 17

\(
F(x) =-x^2+bx+4 = -(x-\alpha )(x-\beta )

\)

Answer 18

comparing x coefficients and constants terms both sides you get :
\(\alpha + \beta = b\)
\(\alpha \beta = -4\)

Answer 19

and you're given another equation :
\( \beta-\alpha = 4\)

Answer 20

3 equations, 3 unknowns - you can solve

Answer 21

thanks @ganeshie8 gladly i didnt screw him with my half knowledge abt this

Answer 22

Practising LaTeX
\[
\begin{align*}
\alpha\beta&=-4\\
\beta&=-\frac{4}{\alpha}
\end{align*}
\\
\]
\[
\begin{align*}
\beta-\alpha&=4\\
-\frac{4}{\alpha}-\alpha&=4\\
-4-\alpha^2&=4\alpha\\
\alpha^2+4\alpha+4&=0\\
(\alpha+2)^2&=0\\
\alpha&=-2\\
\beta&=2\\
b&=0
\end{align*}
\\
\]

Answer 23

haha looks great !! ive never learned these alignment stuff before lol xD

Answer 24

Is it possible to deduce that (0,4) is the vertex given that \(F(x)=-x^2+bx+4\)

Answer 25

definitely, now that u have b value, plug it in and find the vertex

Answer 26

I mean without solving for b. If we can deduce that (0,4) is the vertex, then the roots would clearly be (-2,0) and (2,0) as the x-coordinate of the vertex must lie between the midpoint of the x-coordinates of the roots (by the symmetry of parabola and \(\beta-\alpha=4\)).

Answer 27

Just trying to find shortcuts lol.

Answer 28

I will close this question and repost this so that people can earn more medals. After all, 3 questions and only 1 best medal seems unfair.

Answer 29

il try this later again :)