How do I find the minimum of 5a + b?

  • #1
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Homework Statement


If ax2 - bx + 5 = 0 does not have two distinct real roots, find the minimum value of 5a + b.

2. Homework Equations

The Quadratic Formula

The Attempt at a Solution


Here, D = b2 - 4a(5) = b2 - 20a
D ≤ 0 b2 ≤ 20a b ≤ ±2√(5a)
Also, an obvious observation is that b ≤ 0, since, the quadratic does not have two distinct real roots, all coefficients and the constant are of same sign. So, for a, (-b), 5 to be all positive, b must be non-positive.
But how do I arrive at 5a + b?
 
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Answers and Replies

  • #2
verty
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You've made a mistake. From [itex]b^2 \leq 20a[/itex], one cannot say that [itex]b \leq 2 \sqrt{5a}[/itex].
 
  • #3
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Ok, ±2√(5a).
 
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  • #4
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Also, an obvious observation is that b ≤ 0, since, the quadratic does not have two distinct real roots, all coefficients and the constant are of same sign. So, for a, (-b), 5 to be all positive, b must be non-positive.
(I've now included that in the OP too)
But what next?
 
  • #5
verty
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I think I know how to solve this. The crucial observation is that b^2 = 20a. See if you can explain why.
 
  • #6
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Well, b2 may equal 20a. Or, the maximum value of b2 is 20a.
So, (b2/4)|max = 5a (b/2)2|max = 5a
See if you can explain why.
Because the roots are non-distinct. But how does that seem to give a clue about 5a + b?
 
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  • #7
verty
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Well b has two possible values, right? This is all the help I can give because I don't want go too far with helping.
 
  • #8
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yeah, they are ± 2√(5a).
Now, I'm thinking graphs. The maxima/minima of any quadratic in x occurs at b/2a [since our 'b' is -b] & it's value is -D/4a.
By the way, you know that they're expecting an integral value, right? Not an answer in terms of a or b
 
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  • #9
ehild
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Homework Statement


If ax2 - bx + 5 = 0 does not have two distinct real roots, find the minimum value of 5a + b.

2. Homework Equations

The Quadratic Formula

The Attempt at a Solution


Here, D = b2 - 4a(5) = b2 - 20a
D ≤ 0 b2 ≤ 20a b ≤ ±2√5a
The square root of 20a is not 2√5 a.
Express a with b, and use this relation between a and b in 5a+b.
 
  • #10
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Express a with b, and use this relation between a and b in 5a+b.
Ok, 5a ≥ b2/4 or 5a|min = b2/4
But still, you can only eliminate any one of a & b. We need an integral solution to this.
 
  • #11
ehild
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Ok, 5a ≥ b2/4 or 5a|min = b2/4
But still, you can only eliminate any one of a & b. We need an integral solution to this.
You need to find the minimum of A=5a+b. You know that 5a≥b2/4, that is A=5a+b≥b2/4+b.
What is the minimum value of A?
 
  • #12
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A|min = b2/4+b. That's simply it. How do we get an integral value to that?
 
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  • #13
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Also, an obvious observation is that b ≤ 0, since, the quadratic does not have two distinct real roots, all coefficients and the constant are of same sign.
It might be obvious, but it's incorrect.

The solutions to the equation ##ax^2 - bx + 5 = 0## are ##x = \frac{-b \pm \sqrt{b^2 - 20a}}{2a}## If ##b^2 < 20a##, the roots are complex, and this would include some positive values of b, depending on the value of a.

One thing that hasn't been touched on so far is that the equation "does not have two distinct real roots." This means that there could be one repeated root (Disc = 0) or two complex roots (D < 0).

D ≤ 0 b2 ≤ 20a b ≤ ±2√(5a)
This is wrong as well.
If ##b^2 \le 20a## then ##-2\sqrt{5a} \le b \le 2\sqrt{5a}##. This is different from what you wrote.
 
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  • #14
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Express ##a = a(b) ## or ##b = b(a) ##, then minimise the function ##A(a):= 5a + b(a) ## or ##B(b) := 5a(b) + b##.
 
  • #15
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The crucial observation is that b^2 = 20a.
Not necessarily. The problem statement includes the possibility that the roots are complex, in which case ##b^2 < 20a##.
 
  • #16
Ray Vickson
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Ok, 5a ≥ b2/4 or 5a|min = b2/4
But still, you can only eliminate any one of a & b. We need an integral solution to this.
A|min = b2/4+b. That's simply it. How do we get an integral value to that?

Where in your problem statement was there any requirement that ##a## and ##b## be integers?
 
  • #17
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Where in your problem statement was there any requirement that ##a## and ##b## be integers?
The answer they've given is an integer.
 
  • #18
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The answer they gave is to the question "what is the minimal value of ## 5a+b##". You could have coefficients like ##1.73 ## and ##0.73## or some such.
 
  • #19
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The answer they gave is to the question "what is the minimal value of ## 5a+b##". You could have coefficients like ##1.73 ## and ##0.73## or some such.
No. I think you misunderstood. Their is answer is actually a specific integer (that's one drawback of giving answers. You tend to think in the reverse direction which halts the learning process sometimes).
 
  • #20
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I played around with the problem a bit. Assuming ##b^2 = 20a ## the function of interest is ##B(b) := \frac{b^2}{4} +b ##. If we are to minimise it then the derivative test gives us a candidate for extremum at ##b=-2 ##. The second derivative test confirms it's a minimum point.

Now, one has to explore what happens when the roots can be arbitrary complex numbers.
 
  • #21
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I played around with the problem a bit. Assuming ##b^2 = 20a ## the function of interest is ##B(b) := \frac{b^2}{4} +b ##. If we are to minimise it then the derivative test gives us a candidate for extremum at ##b=-2 ##. The second derivative test confirms it's a minimum point.

Now, one has to explore what happens when the roots can be arbitrary complex numbers.
Wait a minute, are you sure?
I got the extremum value as -1 at b = -2
 
  • #22
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Yes, it was a typo, I fixed it.
 
  • #23
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played around with the problem a bit. Assuming b2=20ab2=20ab^2 = 20a the function of interest is B(b):=b24+bB(b):=b24+bB(b) := \frac{b^2}{4} +b . If we are to minimise it then the derivative test gives us a candidate for extremum at b=−2b=−2b=-2 . The second derivative test confirms it's a minimum point
Can you tell me what thought processes prompted you to do the derivative test?
-1 is the right answer
 
  • #24
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If you are given a function ##f(x) ## that is differentiable and you are asked to find its extrema, then the first thing you do is solve ##f'(x) = 0 ##.

The initial problem is not solved, though. If ##-1## is the right answer, then you have to show ##b^2<20a \implies 5a + b\geq -1 ##.
 
  • #25
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If you are given a function ##f(x) ## that is differentiable and you are asked to find its extrema, then the first thing you do is solve ##f'(x) = 0 ##.
What I mean to say is I get what you did, but I didn't get why.
We already got the function B = b2/4 + b by taking the minimum value of 5a. So why minimise again?
 
  • #26
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@verty largest b given a. b2 equal to 20a or less than that.
 
  • #27
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got the function B = b2/4 + b by taking the minimum value of 5a
underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.
 
  • #28
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underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.
5a ≥ b2/4 right? So the minimum value of 5a is b2/4. Can't neglect that possibility.
Hence, the B ≥ b2/4 + b. Again, B|min = b2/4 + b.
 
  • #29
verty
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I think we need to clear up whether ##a## and ##b## are meant to vary. There was no "given ##a##, ##b##, if ##ax^2 - bx + 5##", it just said "if ##ax^2 - bx + 5##", those variables are unbound, IMHO. But it could be a convention that ##a##,##b## don't vary but ##x##,##y## do.

Ignore this, I was confusing myself.
 
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  • #30
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To elaborate more, B ≥ b2/4 + b is a family of parabolas for all values of b and the 'greater than part' represents the locus of the points outside the interval (α,β) if they're the roots of B
 

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