Quadratic Equation - 3 · KnowledgeBin.org

Now , so k is on left hand side of the point

Also f (k)>0, so it is possible only when k<a.
So,a, b>k.

1) If D³0, f (k)>0 and then both roots of f (x) =0 are less than k. Why?

As D³0 and a>0 so shape of parabola is like this

Fig (9)

Now f (k)>0 so k>b or k<a

And also
So, k<a is ruled out
Therefore k>b
Hence, k>a,b.

2) If D>0 and f (k) <0 then k lies between root of f (x) = 0 why?

D>0 and a>0
So the shape of parabola is

Fig (10)

Now f (k) <0, so clearly a<k<b.

4) If D>0 and f (k1)´f (k2) <0, then exactly one root of equation f (x) = 0 lies in interval (k1, k2). When (k2>k1) Why?

Since f (k1)´f (k2) <0
So, f (k1) and f (k2) should be of opposite sign.

Fig (11)

1)      Now let f(k1)<0 and f(k2)>0
         So, a<k1<b and k2>b (because k2>k1)
         And hence b lies in interval (k1, k2).

2)      Second case is f (k1) >0 and f (k2) <0
         So, a< k2<b and k1<a (because k2>k1)
         Hence a lies in interval (k1, k2).

So exactly one root lies in interval (k1, k2).

5) If D³0, f (p)>0 and f (q)>0, q>p then both the roots of the equation f(x) = 0 will

lie between p and q, if why?

Suppose both roots lie in interval (p, q)

Now D³0, a>0 therefore shape of parabola is

Fig (12)

a, b lies in interval (p, q)

So, b<q and p<a

And hence p<a<b<q

6) If D³0, and then both roots of the equation f(x) = 0 are positive. Why?

Also so, product of root is positive, and hence clearly both roots are positive.

Illustration 8:
For what values of ‘a’ exactly one root of the equation lies between 1 and 2.

Solution: Since exactly one root of given equation lie between 1 and 2,
So f (1)´f (2) <0
Here f(x) =
So, f (1)´f (2) <0

Some important results using differentiability:

1) If f(x) = 0 has a real root a of multiplicity g (g>1) then f(x) = (x-a)gg (a) where g(a)¹0. Also f(x) = 0 has a as a real root with multiplicity (g-1)

2) If f(x) = 0 has n real roots then f (x) = 0 has (n-1) real roots.

Illustration 9:
A polynomial p(x) has as one of the factors, other roots of this polynomial lie in the range Prove that g(x) where g(x) = p1(x) has at least one positive root.

Solution:

Now p(x) has a root 2 with multiplicity 2.
So the derivative of p(x), p1(x) or g(x) must has 2 as root with multiplicity 2-1=1.
So g(x) has at least one positive root which is 2.