Hi Guys, these are some questions I really need help with,
1. The quadratic equation ax^2+bx+1=0 has one root twice the other. Prove that 2b^2=9a
2. Show that y>0, for all x, for y=x^2+2x+4
Thanks :D
1. The quadratic equation ax^2+bx+1=0 has one root twice the other. Prove that 2b^2=9a
2. Show that y>0, for all x, for y=x^2+2x+4
Thanks :D
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Let be m and 2m the roots of ax^2+bx+1=0 then
x^2+(b/a)x + 1/a = 0
(x - m)(x-2m) = 0
x^2 - 3mx +2m^2 = 0
b/a = 3m
1/a = 2m^2
1/a = 2* (b/3a)^2
2b^2 = 9a
Show that y>0, for all x, for y=x^2+2x+4
y = (x+1)^2 + 3 > or equal to zero for any real x because is sum of perfect square
x^2+(b/a)x + 1/a = 0
(x - m)(x-2m) = 0
x^2 - 3mx +2m^2 = 0
b/a = 3m
1/a = 2m^2
1/a = 2* (b/3a)^2
2b^2 = 9a
Show that y>0, for all x, for y=x^2+2x+4
y = (x+1)^2 + 3 > or equal to zero for any real x because is sum of perfect square
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Not sure how to answer question 1, but I can answer #2.
All of the values in #2 are positive. a positive x^2, a positive x and a positive constant.
The positive constant and x values mean that the vertex is above the x-axis (if you aren't sure, make it into vertex form. I would suggest doing this as proof anyways.)
The positive x^2 means that the parabola opens up.
Since it opens up and the lowest point (vertex) is above the x-axis, it will never go below the x-axis and y will always be greater than 0.
All of the values in #2 are positive. a positive x^2, a positive x and a positive constant.
The positive constant and x values mean that the vertex is above the x-axis (if you aren't sure, make it into vertex form. I would suggest doing this as proof anyways.)
The positive x^2 means that the parabola opens up.
Since it opens up and the lowest point (vertex) is above the x-axis, it will never go below the x-axis and y will always be greater than 0.
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Use the quadratic formula:
roots are at
x = [ -b +/- sqrt( b^2 - 4ac ) ] / 2a
Let us suppose that the roots are positive, and that, therefore, the one with the + sqrt will be twice the value of the one with the - sqrt:
[ -b + sqrt( b^2 - 4ac ) ] / 2a = 2 [ -b - sqrt( b^2 - 4ac ) ] / 2a
Multiply both sides by 2a (gets rid of the denominator):
[ -b + sqrt( b^2 - 4ac ) ] = 2 [ -b - sqrt( b^2 - 4ac ) ]
distribute the 2 in the square bracket (right side)
[ -b + sqrt( b^2 - 4ac ) ] = [ -2b - 2sqrt( b^2 - 4ac ) ]
get rid of the square brackets (they no longer serve any purpose):
-b + sqrt( b^2 - 4ac) = -2b -2sqrt(b^2 - 4ac)
move -2sqrt to the left and -b to the right
sqrt(b^2-4ac) + 2sqrt(b^2 - 4ac) = -2b + b
3sqrt(b^2 - 4ac) = -b
square both sides (gets rid of the sqrt)
9(b^2 - 4ac) = b^2
8b^2 = 36ac
2b^2 = 9ac
roots are at
x = [ -b +/- sqrt( b^2 - 4ac ) ] / 2a
Let us suppose that the roots are positive, and that, therefore, the one with the + sqrt will be twice the value of the one with the - sqrt:
[ -b + sqrt( b^2 - 4ac ) ] / 2a = 2 [ -b - sqrt( b^2 - 4ac ) ] / 2a
Multiply both sides by 2a (gets rid of the denominator):
[ -b + sqrt( b^2 - 4ac ) ] = 2 [ -b - sqrt( b^2 - 4ac ) ]
distribute the 2 in the square bracket (right side)
[ -b + sqrt( b^2 - 4ac ) ] = [ -2b - 2sqrt( b^2 - 4ac ) ]
get rid of the square brackets (they no longer serve any purpose):
-b + sqrt( b^2 - 4ac) = -2b -2sqrt(b^2 - 4ac)
move -2sqrt to the left and -b to the right
sqrt(b^2-4ac) + 2sqrt(b^2 - 4ac) = -2b + b
3sqrt(b^2 - 4ac) = -b
square both sides (gets rid of the sqrt)
9(b^2 - 4ac) = b^2
8b^2 = 36ac
2b^2 = 9ac
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2. Show that y > 0, for all x, for y = x^2 + 2x + 4
First off,
y = x^2 + 2x + 4 implies that
y = x^2 + 2x + 1 + 3
y = (x + 1)^2 + 3
Since (x + 1)^2 is definitely greater than or equal to 0, that means
(x + 1)^2 >= 0
so if we add 3 both sides,
(x + 1)^2 + 3 >= 3
and 3 is strictly greater than 0, so
(x + 1)^2 + 3 > 0, i.e.
y > 0
First off,
y = x^2 + 2x + 4 implies that
y = x^2 + 2x + 1 + 3
y = (x + 1)^2 + 3
Since (x + 1)^2 is definitely greater than or equal to 0, that means
(x + 1)^2 >= 0
so if we add 3 both sides,
(x + 1)^2 + 3 >= 3
and 3 is strictly greater than 0, so
(x + 1)^2 + 3 > 0, i.e.
y > 0