Question

Help with finding the roots.
(Image Included)

Answer 1

@alivejeremy do you mind helping me ?

Answer 2

SOrry i'm busy at the moment

Answer 3

kk

Answer 4

here we have to write the equation of such parabola, first

Answer 5

the equation which models the parabola, is:
\(y=ax^2+bx+c\)

Answer 6

and, using the points you provided, we have to determine the values of the coefficients \(a,b,c\)

Answer 7

kk

Answer 8

for example I consider the point \((0,4)\)
after the substitution into the equation above, we get:
\[\begin{gathered}
4 = a \cdot {0^2} + b \cdot 0 + c \hfill \\
\hfill \\
c = 4 \hfill \\
\end{gathered} \]

Answer 9

kk and the next time is to ?

Answer 10

next I consider the point \((2,0)\), and, again, after a substitution, I get:
\[\begin{gathered}
0 = a \cdot {2^2} + b \cdot 2 + c \hfill \\
\hfill \\
0 = 4a + 2b + c \hfill \\
\end{gathered} \]

Answer 11

finally, I consider the point \((4,0)\), and in a similar way, I can write:
\[\begin{gathered}
0 = a \cdot {4^2} + b \cdot 4 + c \hfill \\
\hfill \\
0 = 16a + 4b + c \hfill \\
\end{gathered} \]

Answer 12

so, we got these three conditions:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
0 = 4a + 2b + c \hfill \\
0 = 16a + 4b + c \hfill \\
\end{gathered} \right.\]
which is a linear system for \(a,b,c\), please solve for \(a,b,c\)

Answer 13

please wait a moment, I think that the point \((4,0)\) is not correct! Since the parabola does \(not\) pass at such point

Answer 14

alrighty

Answer 15

oh yeah i made an error

Answer 16

it should be -4,0

Answer 17

ok! In this case the last equation has to be replaced with this one:
\[0 = 16a - 4b + c\]
so, the final linear system that we have to solve is:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
0 = 4a + 2b + c \hfill \\
0 = 16a - 4b + c \hfill \\
\end{gathered} \right.\]

Answer 18

please try to solve such system

Answer 19

a=0 b=0

Answer 20

hint:
I substitute \(c=4\) into the second and third equation, so I get:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
0 = 4a + 2b + 4 \hfill \\
0 = 16a - 4b + 4 \hfill \\
\end{gathered} \right.\]

Answer 21

ughhh i forgot about substituting it

Answer 22

next I divide the second equation by 2, and the third equation by 4, so I write this:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
0 = 2a + b + 2 \hfill \\
0 = 4a - b + 1 \hfill \\
\end{gathered} \right.\]

Answer 23

I solve the second equation for \(b\), and I get:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
b = - 2a - 2 \hfill \\
0 = 4a - b + 1 \hfill \\
\end{gathered} \right.\]

Answer 24

now, I substitute the value of \(b\) into the third equation:
\[\left\{ \begin{gathered}
c = 4 \hfill \\
b = - 2a - 2 \hfill \\
0 = 4a - \left( { - 2a - 2} \right) + 1 \hfill \\
\end{gathered} \right.\]
please solve the third equation for \(a\)

Answer 25

a=\[\frac{ -1 }{ 2}\]

Answer 26

correct!
now, the value of \(b\) is:
\[b = - 2 \cdot \left( {\frac{{ - 1}}{2}} \right) - 2 = ...?\]

Answer 27

22

Answer 28

hint:
we have \(-2 \cdot (- 1/2)=1\), so:
\[b = - 2 \cdot \left( {\frac{{ - 1}}{2}} \right) - 2 = 1 - 2 = ...?\]

Answer 29

-1

Answer 30

I'm confused

Answer 31

correct!
so the cartesian equation of the parabola is:
\[y = \frac{{ - {x^2}}}{2} - x + 4\]

Answer 32

next, the symmetry axis has the subsequent equation:
\[\Large x = - \frac{b}{{2a}} = - \frac{{ - 1}}{{2 \cdot \left( {\frac{{ - 1}}{2}} \right)}} = ...?\]

Answer 33

please complete

Answer 34

hint:
\[\Large \begin{gathered}
x = - \frac{b}{{2a}} = - \frac{{ - 1}}{{2 \cdot \left( {\frac{{ - 1}}{2}} \right)}} = \hfill \\
\hfill \\
= \frac{1}{{2 \cdot \left( {\frac{{ - 1}}{2}} \right)}} = \frac{1}{{ - 1}} = ...? \hfill \\
\end{gathered} \]

Answer 35

I think it is \(x=-1\)
am I right?

Answer 36

yeah that's what I got as well.........

Answer 37

ok! So the x-coordinate of the vertex is \(x=-1\) and \(not\) x=0
furthermore, by substitution, the y-coordinate of the vertex, is:
\[{y_V} = \frac{{ - {{\left( { - 1} \right)}^2}}}{2} - \left( { - 1} \right) + 4 = ...?\]

Answer 38

lol yea typoss

Answer 39

oops.. I meant and not x=1*

Answer 40

please evaluate \(y_V\)
\[{y_V} = \frac{{ - {{\left( { - 1} \right)}^2}}}{2} - \left( { - 1} \right) + 4 = ...?\]

Answer 41

27 ?

Answer 42

hint:
\[\begin{gathered}
{y_V} = \frac{{ - {{\left( { - 1} \right)}^2}}}{2} - \left( { - 1} \right) + 4 = \frac{{ - 1}}{2} + 1 + 4 = \hfill \\
\hfill \\
= \frac{{ - 1}}{2} + 5 = ...? \hfill \\
\end{gathered} \]

Answer 43

4.5 or 9/2

Answer 44

correct! So the vertex, is:
\(V=(-1,\;9/2)\)

Answer 45

finally the roots can be computed solving this quadratic equation:
\[\frac{{ - {x^2}}}{2} - x + 4 = 0\]
please try to solve such equation

Answer 46

so i used the calculator and i got -1/2 + 1/2\[\sqrt{17}\]

Answer 47

hint:
if I multiply such equation by \(-2\), I get this:
\[{x^2} + 2x - 8 = 0\]

Answer 48

now, from the general theory on quadratic equations, we can write this:
\[x = \frac{{ - 2 \pm \sqrt {4 + 32} }}{2} = \frac{{ - 2 \pm \sqrt {36} }}{2} = ...?\]
please continue

Answer 49

ugh this is really confusing and I'm not understanding it... I give up but thanks for your help @Michele_Laino :)

Answer 50

if we start from this equation:
\[{x^2} + 2x - 8 = 0\]
I have to apply this formula:
\[x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
wherein \(a=1,b=2,c=-8\), so, after a substitution, I get:
\[x = \frac{{ - 2 \pm \sqrt {4 + 32} }}{2} = \frac{{ - 2 \pm \sqrt {36} }}{2} = ...?\]

Answer 51

here is the next step:
\[\begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \hfill \\
x = \frac{{ - 2 \pm \sqrt {4 + 32} }}{2} = \frac{{ - 2 \pm \sqrt {36} }}{2} = \frac{{ - 2 \pm 6}}{2} \hfill \\
\end{gathered} \]
so:
\[\begin{gathered}
{x_1} = \frac{{ - 2 + 6}}{2} = ...? \hfill \\
\hfill \\
{x_2} = \frac{{ - 2 - 6}}{2} = ...? \hfill \\
\end{gathered} \]

Answer 52

that's equal to -4

Answer 53

yes, the first root is \(x_1=2\), whereas the second root is \(x_2=-4\)

Answer 54

ahh ok ok I see omg !!!!!! thank you so MUCH !!!!! @Michele_Laino you're a genius !!!!

Answer 55

thanks! :) :)

Answer 56

lol you're welcome

Answer 57

thanks! :) :) :)