Answer:First suppose that the roots of the equation
x2−bx+c=0(1)
are real and positive. From the quadratic formula, we see that the roots of (1) are of the form
b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2.
For the root or roots to be real, we require that b2−4c≥0, that is, b2≥4c. In order for them to be positive, we require that
b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0.
This immediately tells us that b>0, but we can go further. We can rearrange this to get
b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,
which (assuming that b>0) is true if and only if
b2>b2−4c,
since both sides of the inequality are positive so we may square. But then
4c>0.
That is, if the roots are real and positive then b>0 and b2≥4c>0.
Now suppose that b>0 and b2≥4c>0.
Then the roots of (1) are real since b2−4c≥0, and b>0 guarantees that the root b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ is positive.
So it remains to show that b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0. We have that
4c>0,
so that
b2>b2−4c,
then square rooting shows that
b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,
so the roots of (1) are real and positive, as required.
Approach 2
the curve y equals x squared minus b x plus c showing two positive roots for y equals zero
This is intended to be a proof without words! We have from the diagram that:
If c, b and b2−4c are all positive, there are two real positive roots for x2−bx+c=0 (if b2=4c, we have two real positive equal roots).
If there are two real positive roots for x2−bx+c=0, then c and b are positive and b2−4c is non-negative.
(Why is the distance between the roots b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√?)
Approach 3
When solving problems about the roots of polynomials, it is often useful to find expressions those roots must satisfy and see if this tells us anything new. If α and β denote the roots of the equation, then
x2−bx+c=(x−α)(x−β)=x2−(α+β)x+αβ
and so α+β=b and αβ=c.
We also know that the roots of a quadratic equation are real if and only if the discriminant is non-negative, that is, if and only if b2−4c≥0.
Using these facts, if α and β are both real and positive, then b=α+β>0, c=αβ>0 and b2≥4c, as above.
Conversely, if b>0 and b2≥4c>0, then we know the discriminant is positive and hence both roots are real. We also have that
αβ>0(2)
and
α+β>0.(3)
As α and β are both real, by (2), we know that α and β are either both positive or both negative. However, if α and β were both negative, then (3) could not possibly hold. Hence α and β are both positive, as required.
We now sketch on a graph the region where b>0, c>0 and b2≥4c:
The curve b squared = 4 c as a quadratic with the c-axis vertical and the b-axis horizontal. The region below it is shaded.
The region of the b-c plane for which b>0, c>0 and b2≥4c
Sketch the region of the b-c plane in which the roots of the equation are real and less than 1 in magnitude.
We know that in order for the roots to be real we need b2≥4c as in the first part. We now need to find the region where
−1<b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.(4)
We have b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2≤b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2 so we only need to consider the values for which both
−1<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2andb+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.
Firstly, we will consider the values for which
b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2.
Rearranging gives
b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2−b.
So b<2, as the square root is non-negative, and we can square both sides to get
b2−4c<4−4b+b2,
which we may rearrange to find c>b−1.
We will now consider the values for which
−2<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√.
Similarly, we can rearrange to get
b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<b+2.
So b>−2, and we can, as before, square to get
b2−4c<b2+4b+4,
and hence c>−b−1.
To sketch the graph, we start by considering the boundary curves b2=4c, c=b−1 and c=−b−1, and the points at which they intersect. We can see that the two lines only intersect when b=0 and c=−1, and the lines intersect the curve when
b2=4b−4andb2=−4b−4
which rearrange to
(b−2)2=0and(b+2)2=0.
This tells us that each line intersects with the curve in only one place and so these lines must be tangent.
Is there a way we could have deduced this directly from (4)?
What do the lines being tangent signify in terms of our equation x2−bx+c=0?
Is this a representation of a well-known property of these equations?
Sketching the graph then yields the following picture:
The graph with the previous curve and the lines 4 c = 4 b minus 4 and 4 c = minus 4 b - 4. Each line touches the curve once and the two lines intersect at (0, minus 1). The region between the three lines/curves is shaded.
The shaded region is where b2≥4c, c>b−1, c>−b−1 and −2<b<2
We might notice that when we derived the inequalities c>b−1 and c>−b−1, a lot of the work we did was very similar.
We might also notice that the graph above is symmetric about the c axis. Why is this?
Does the graph give us any ideas about how we might deduce one inequality from the other?
Step-by-step explanation:
