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28 Aug 2019, 00:00
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I am bit confused with quadratic equation.please help Bunuel Kinshook @e-GmatQuant nick1816
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
28 Aug 2019, 10:46
Kudos
Bookmarks
if the roots of equation quadratic equation are -1 and 3, then quadratic equation must be in the form of a(x+1)(x-3)
x co-ordinates of vertices will always be 1, but y co-ordinates of vertices will depend on value of a.
When a=-1, Vertices of parabola = (1, 4)
When a=-2, Vertices of parabola = (1, 8)
..... so on
Any curve of form, a(x+1)(x-3), will always cut x axis at -1 and 3. Value of 'a' only affects y co-ordinates of vertices.
x co-ordinates of vertices will always be 1, but y co-ordinates of vertices will depend on value of a.
When a=-1, Vertices of parabola = (1, 4)
When a=-2, Vertices of parabola = (1, 8)
..... so on
Any curve of form, a(x+1)(x-3), will always cut x axis at -1 and 3. Value of 'a' only affects y co-ordinates of vertices.
vanam52923
I am bit confused with quadratic equation.please help [url=https://gmatclub.com:443/forum/memberlist.php?
mode=viewprofile&un=Bunuel]Bunuel[/url] Kinshook @e-GmatQuant nick1816
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
mode=viewprofile&un=Bunuel]Bunuel[/url] Kinshook @e-GmatQuant nick1816
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
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28 Aug 2019, 18:29
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nick1816
if the roots of equation quadratic equation are -1 and 3, then quadratic equation must be in the form of a(x+1)(x-3)
x co-ordinates of vertices will always be 1, but y co-ordinates of vertices will depend on value of a.
When a=-1, Vertices of parabola = (1, 4)
When a=-2, Vertices of parabola = (1, 8)
..... so on
Any curve of form, a(x+1)(x-3), will always cut x axis at -1 and 3. Value of 'a' only affects y co-ordinates of vertices.
ty so much x co-ordinates of vertices will always be 1, but y co-ordinates of vertices will depend on value of a.
When a=-1, Vertices of parabola = (1, 4)
When a=-2, Vertices of parabola = (1, 8)
..... so on
Any curve of form, a(x+1)(x-3), will always cut x axis at -1 and 3. Value of 'a' only affects y co-ordinates of vertices.
vanam52923
I am bit confused with quadratic equation.please help [url=https://gmatclub.com:443/forum/memberlist.php?
mode=viewprofile&un=Bunuel]Bunuel[/url] Kinshook @e-GmatQuant nick1816
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
mode=viewprofile&un=Bunuel]Bunuel[/url] Kinshook @e-GmatQuant nick1816
VeritasKarishma
if the roots of equation quadratic equation are -1 and 3
so we can write it in the form (x+1)(x-3)=0
Also, we know x^2-sum of roots *x+product of roots=0
x^2 -(2)x-3=0
Now my question is
if it is given -1 and 3 are roots of ax^2 + bx + c=0
can we write it as (x+1)(x-3)=0
or will it be a(x+1)(x-3)=0,if this then how?
Here the questioin which confused me was :
y= ax^2 + bx + c
The equation above represents a parabola in the XY-plane. If a < 0 and the x-intercepts of the parabola are – 1 and 3, which of the following could be the vertex of the parabola?
so if for x intercept i put y=0 then i get
ax^2 + bx + c=0 also i knw -1 and 3 are roots then it will become
(x+1)(x-3)=0
(x^2 -2x -3)
but i need a(x^2 -2x -3) bcz a <0 given then only i will come to know
ax^2 - 2ax -3a = a{(x-1)^2 -4}
For vertex of the parabola
x = 1
y = -4a since a<0
y = 4k where k>0
please clear
