Unit 1/2 Problem

[clarke54321]

Hi All,

Could I please have some help/explanations for the following discriminant problems.

1) Find the discriminant of the equation 4x^2 + (m-4)x -m= 0, where m is a rational number, and hence show that the equation has rational solution(s).

2) If both a and c are positive, what can be said about the graph of y= ax^2 + bx + c


Thanks in advance 

[keltingmeith]

1. The discriminant is b^2-4ac=(m-4)^2-4(4)(-m)=m^2-8m+16+16m=m^2+8m+16=(m+4)^2

This number is always greater than or equal to 0 (consider the graph of y=(m+4)^2, where y is graphed against m), and so real solutions always exist. Proving that these solutions are rational is not easy, but remember that this value will be under the square-root of the quadratic equation. The sum of two rational numbers is, again, rational, and the square root of the square returns a rational number, ergo you'll get a rational solution.

I believe the rational bit is waaaaay beyond 1/2 level, probably even 3/4 level. You could be asked to show that this quadratic always has real solutions, but I reckon rational is a bit far, so don't stress too much if this confused you.

2. This one is also quite difficult! I'd suggest to again consider the discriminant, and see what you can gleam from that. In truth, for this question, there are LOTS of different ways you could answer it, because it's so open. Just see how many different answers you can come up with!

[clarke54321]

1. The discriminant is b^2-4ac=(m-4)^2-4(4)(-m)=m^2-8m+16+16m=m^2+8m+16=(m+4)^2

This number is always greater than or equal to 0 (consider the graph of y=(m+4)^2, where y is graphed against m), and so real solutions always exist. Proving that these solutions are rational is not easy, but remember that this value will be under the square-root of the quadratic equation. The sum of two rational numbers is, again, rational, and the square root of the square returns a rational number, ergo you'll get a rational solution.

I believe the rational bit is waaaaay beyond 1/2 level, probably even 3/4 level. You could be asked to show that this quadratic always has real solutions, but I reckon rational is a bit far, so don't stress too much if this confused you.

2. This one is also quite difficult! I'd suggest to again consider the discriminant, and see what you can gleam from that. In truth, for this question, there are LOTS of different ways you could answer it, because it's so open. Just see how many different answers you can come up with!

Thanks Euler, this is a great help.

For the second question, my text book says 'the graph will cross the x-axis twice' meaning the discriminant is greater than 0. But how am I supposed to show this?

[keltingmeith]

Thanks Euler, this is a great help.

For the second question, my text book says 'the graph will cross the x-axis twice' meaning the discriminant is greater than 0. But how am I supposed to show this?

... lolwut. This is definitely not true.

We have the quadratic f(x)=ax^2+bx+c, with a,c>0 (that is, both a and c are positive). No stipulations on b.

So, the discriminant is b^2-4ac (literally). If there are no solutions, then we require b^2<4ac

So, if we choose b, a and c so that this is satisifed, we have no solutions. So, I'm going to pick b=0, a=c=1, so we have f(x)=x^2+1. This has no solutions and so does not cross the x-axis at all. Satisfies everything they said, but goes against their answer.

[clarke54321]

... lolwut. This is definitely not true.

We have the quadratic f(x)=ax^2+bx+c, with a,c>0 (that is, both a and c are positive). No stipulations on b.

So, the discriminant is b^2-4ac (literally). If there are no solutions, then we require b^2<4ac

So, if we choose b, a and c so that this is satisifed, we have no solutions. So, I'm going to pick b=0, a=c=1, so we have f(x)=x^2+1. This has no solutions and so does not cross the x-axis at all. Satisfies everything they said, but goes against their answer.

Thanks for the clarification 

[clarke54321]

Hi everyone,

 Can someone please help me with the following question:

Find the value of m for which the pair of simultaneous equations 3x + my = 5 and (m + 2)x + 5y = m have:

a) infinitely many solutions
b) no solutions.


Thank-you 

[Peanut Butter]

Hi everyone,

 Can someone please help me with the following question:

Find the value of m for which the pair of simultaneous equations 3x + my = 5 and (m + 2)x + 5y = m have:

a) infinitely many solutions
b) no solutions.


Thank-you 

Infinitely many solutions: m1 = m2 and c1 = c2

No solutions: m1 = m2 and c1 does not equal c2

First step, rearrange both the equations in terms of y.

(1) y = -3x/m + 5/m
(2) y = -(m+2)x/5 + m/5

Make m1 = m2, and then solve for m.

-3/m = -(m+2)/5

m^2 + 2m - 15 = 0

(m + 5)(m-3) = 0

Therefore, m = -5 or m = 3

Sub in these values to find the y intercept. Where for infinite solutions c1 = c2, and no solutions c1 does not equal c2.

a) infinitely many solutions: m = -5 (because 5/-5 = -5/5)
b) no solutions: m = 3 (because 5/3 does not equal 3/5)

Hope this helped!

[clarke54321]

Infinitely many solutions: m1 = m2 and c1 = c2

No solutions: m1 = m2 and c1 does not equal c2

First step, rearrange both the equations in terms of y.

(1) y = -3x/m + 5/m
(2) y = -(m+2)x/5 + m/5

Make m1 = m2, and then solve for m.

-3/m = -(m+2)/5

m^2 + 2m - 15 = 0

(m + 5)(m-3) = 0

Therefore, m = -5 or m = 3

Sub in these values to find the y intercept. Where for infinite solutions c1 = c2, and no solutions c1 does not equal c2.

a) infinitely many solutions: m = -5 (because 5/-5 = -5/5)
b) no solutions: m = 3 (because 5/3 does not equal 3/5)

Hope this helped!

Thanks Peanut Butter!