Quadratic Function/roots Queries! (1 Viewer)

[Finx]

Just a bit lost on a few doozies;

2. Find the equation of a quadratic function that passes through the points (-2,18), (3,-2) and (1,0).
I know that these points are the roots for the quadratic function, but I'm not sure how to work it out =[

3. Find the value of a, b and c if
x² + 5x - 3 = ax(x+1) + b(x+1)² + cx
No, I didn't make any typos. I was thinking of subbing X for (x+1), which means I yield axX + bX + cx, but I still can't see how it finds a, b and c.

6. Find the value of n for which the equation (n+2)x² + 3x - 5 = 0 has one root triple the other.
I have a very rough idea of how this one might be worked out, but I'd like to see how it's done properly.

Thanks in advance!

 

[e-star!09]

err, i can't think of how to do the first question right now, but i'm sure someone else will figure it out ^^

x² + 5x - 3 = ax(x+1) + b(x+1)² + cx
x² + 5x - 3 = ax² + ax + bx² + 2bx + b + cx
x² + 5x - 3 = (a + b)x² + (2b + c)x + b
so, that means,
a + b = 1___ eq.1
2b +c = 5___eq.2
b = -3
sub b= -3 into eq.1, --> a - 3 = 1 --> a= 4
sub b= -3 into eq.2 --> 2(-3) + c = 5 --> -6 + c = 5 --> c = 11
therefore, a= 4, b= -3, c= 11

6. (n + 2)x² + 3x - 5 = 0
a= n + 2, b= 3, c= -5
let the roots be β and 3β
β + 3β = -b/a
4β = -3/(n + 2)
β = -3/[4(n + 2)]___ eq.1
β x 3β = c/a
3β² = -5/(n + 2)___ eq.2
sub eq.1 into eq.2
3{-3/[4(n + 2)]}² = -5/(n + 2)
27/[16(n + 2)²] = -5/(n + 2)
27(n + 2) = -5[16(n + 2)²]
27n + 54 = -5(16n² + 64n + 64)
27n + 54 = -80n² + 320n + 320
80n² - 293n - 266 = 0
and.... u could probably just solve the rest...
*yawns* sleepy...

 

[M@ster P]

ok the first question asks you to find the quadratic function or equation.

The general form of a quadratic function is given by y = ax^2 + bx + c. The question also gives you three points (-2,18), (3,-2) and (1,0).

Now using the general form of the quadratic equation, sub each of the 3 points to form 3 different equations.

using (-2, 18) we get:

18 = a(-2)^2 - 2b + c
18 = 4a - 2b + c

This gives you one equation using the point (-2, 18). Now do the exact same thing for the other two points (3,-2) and (1,0) and you will get 2 more equations. In total you will get 3 different equations from the three points that were given to you.

The next step is to solve the 3 equations simultaneously to find the respective values of a, b and c. Once you find these values you can finally form the quadratic function by subbing the values of a, b and c into the general form of a quadratic equation y = ax^2 + bx + c

Hope you understand.
​

 

[youngminii]

The next step is to solve the 3 equations simultaneously to find the respective values of a, b and c. Once you find these values you can finally form the quadratic function by subbing the values of a, b and c into the general form of a quadratic equation y = ax^2 + bx + c

Hope you understand.

​

Just completing it 'cause I'm bored~
y = ax^2 + bx + c, where a =/= 0

Subbing in (-2, 18)

18 = 4a - 2b + c
c = 18 - 4a + 2b ----> 1.

Subbing in (3, -2)

-2 = 9a + 3b + c
3b = -2 - 9a - c ----> 2.

Subbing in (1, 0)

0 = a + b + c
a = -b - c ----> 3.

Sub 3. -> 2.

3b = -2 - 9(-b - c) - c
3b = -2 + 9b + 9c - c
c = (-3b - 1)/4 ----> 4.

Sub c and a -> 1.

(-3b - 1)/4 = 18 + 4(-3b - 1)/4 + 4b + 2b
-3b - 1 = 72 -12b -4 + 16b + 8b
-1 - 72 + 4 = -12b + 16b + 8b + 3b
-69 = 15b
b = -69/15
= -23/5

Sub b -> 4.
c = (-3(-69/15)-1)/4
= 192/15/4
= 192/60
= 16/5

Sub b and c -> 3.
a = -((-69/15) + (192/60))
= 69/15 - 192/60
= 276/60 - 192/60
= 84/60
= 7/5

Subbing a, b and c into ax^2 + bx + c, where a =/= 0
7x^2 - 23x + 16 = 0

I am probably wrong 'cause those numbers look ugly

 

[Zeber]

Just a bit lost on a few doozies;

2. Find the equation of a quadratic function that passes through the points (-2,18), (3,-2) and (1,0).
I know that these points are the roots for the quadratic function, but I'm not sure how to work it out =[

The quadratic function will be in the form ax^2 + bx + c
(1,0) is when x = 1, y = 0
(1) a+b+c = 0
x = -2
(2) 4a -2b + c = 18
x = 3
(3) 9a + 3b + c = -2
solve simulatenously.

3. Find the value of a, b and c if
x² + 5x - 3 = ax(x+1) + b(x+1)² + cx
No, I didn't make any typos. I was thinking of subbing X for (x+1), which means I yield axX + bX + cx, but I still can't see how it finds a, b and c.

x^2 + 5x -3 = ax^2 + ax + b(x^2+2x+1) + cx
RHS = ax^2 + ax + bx^2 + b2x + b + cx
= (a+b)x^2 + (a+2b+c)x + b

Equation coefficients.

b = -3
a- 3 = 1
a = 4

4-6 + c = 5
-2 + c = 5
c = 7

6. Find the value of n for which the equation (n+2)x² + 3x - 5 = 0 has one root triple the other.

by letting roots a, 3a.

3a+a = -3/n+2 = 4a (1)
3a^2 = -5/n+2 (2)

from (1), a = -3/4(n+2)

sub into (2) and solve.

 

[Finx]

Thanks guys, its all worked out now ^_^.

 

[DiligentStudent]

Lol a HSC 2010 student solved your simple questions. Fucken pathetic. Why are you even doing 4U? Epic fail.

 

[Finx]

Lol a HSC 2010 student solved your simple questions. Fucken pathetic. Why are you even doing 4U? Epic fail.

Zeber's profile says he's a 2011 HSC student. This means he would be in Year 9 at the moment. Do you really think someone in year 9 would have the knowledge to do these questions? I doubt Zeber is actually a 2011'er.

Trolls are cool. True fact.

 

[Zeber]

Zeber's profile says he's a 2011 HSC student. This means he would be in Year 9 at the moment. Do you really think someone in year 9 would have the knowledge to do these questions? I doubt Zeber is actually a 2011'er.

Trolls are cool. True fact.

a year 9 with a year 12 text book would have the knowledge.

 

[Finx]

a year 9 with a year 12 text book would have the knowledge.

Are you really y9? o_o

 

[Zeber]

Are you really y9? o_o

yes. my dad is a maths teacher hence he teaches me.

 

[Finx]

Nice..

 

[Timothy.Siu]

Zeber's profile says he's a 2011 HSC student. This means he would be in Year 9 at the moment. Do you really think someone in year 9 would have the knowledge to do these questions? I doubt Zeber is actually a 2011'er.

yes. my dad is a maths teacher hence he teaches me.

owned