Need help with polynomials?

Question

1) Express f(x)=x²-4x+8 in the form (x-p)²+q. Deduce that
f(x)>0 for all x E IR.

2) Find the set of values of a where ax²+3x+4 is positive for all real values of x.

3) Show that the expression -x²+px-q is negative for all real values of x if q > 1/4 p²

Answer 1

1) f(x)=(x-2)^2+4 ..................(x-p)^2+q
now f(x)>0 ,only if (x-2)^2 > 0
we know that square of any number (positive or negative) is always positive
therefore (x-2)^2 >=0 (greater than or equal to)
therefore given f(x) > 0 .......for all x € IR

2) f(x) = a*( x^2+(3/a)*x ) + 4 ;....taking 'a' common
= a*(x+1.5/a)^2 - a*(1.5/a)^2 + 4
= a*(x+1.5/a)^2 - 2.25/a + 4
first term will be greater than zero for all 'a' >0 (x € IR)
the sum (-2.25/a + 4) will also be greater than zero for all 'a' except a=0
therefore for a>0 f(x) will be positive

3) let f(x) = -(x^2-px+q)
= -[ (x-p/2)^2 -1/4p^2 + q ]
= [ -(x-p/2)^2 ] + [ -(q - 1/4p^2) ]
first term will be negative for all 'x' and 'p'
second term will be negative iff (q - 1/4p^2) >0
i.e. q > 1/4 p²

Answer 2

The first question has two parts. The first part is a form of "completing the square".

x^2-4x+8 is not a perfect square. However, it is possible to write if as a square plus a remainder.

For example:
x^2-4x+8 = x^2-4x+4+4 (changed the +8 to +4+4)
The perfect square portion is x^2-4x+4 which is equal to
(x-2)^2

Therefore:
x^2-4x+8 = x^2-4x+4+4 = (x-2)^2 + 4

The second part is a simple analysis: In real numbers, no square can be negative. A square is either 0 (the square of 0) or it is positive (the square of any number other than zero).

Therefore the value of f(x) is the sum of a square (x-2)^2 which is never negative, PLUS the value +4, which is always positive.

The value of f(x) can never be less than +4.
(That, by itself, is sufficient proof that f(x) has no real roots)

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We want
ax^2 + 3x +4 >= 0 (greater than or equal to)
At x=0, this is true (+4 >= 0)

For x not equal to zero

ax^2 >= -3x - 4
a >= - 3/x - 4/x^2

This is OK: we divided by x^2 which cannot be zero (because we set x not equal to zero) and x^2 is always positive (it is a square) so that we do not have to worry about the direction of the inequality -- we have divided by a positive value).

a >= -3/x - 4/x^2

-4/x^2 will always be negative. Therefore, a>=0 will always be bigger than -4/x^2.

a>=0 will always be bigger than -3/x when x is positive.

So we only need to worry about negative x.

what is the maximum value of g(x) = -3/x -4/x^2

It will be wherever g'(x) = 0 = 3/x^2 + 8/x^3

0 = 3x/x^3 + 8/x^3

which will occur when 0 = 3x + 8
x = -8/3

g(-8/3) = -3/(-8/3) -4/(64/9)
= 9/8 - 36/64
= 72/64 - 36/64 = 36/64 = 9/16

So, a must be greater or equal to +9/16.

Answer 3

1). Take half of 4 and square it to complete the
square. We can write the expression as
x²-4x+4+4 = (x-2)²+4.
Since (x-2)² >= 0, we conclude f(x)>0 for all real x.

2). The simplest way to do this one is to compute
the discriminant. If ax²+3x+4 is positive for all real x
its graph never crosses the x axis, so its discriminant
must be negative.
So 9 -16a<0
16a > 9
a > 9/16.

3). If -x²+px-q is negative, x²-px+q is positive.
So, when is the latter positive for all real x?
The same idea as in #2 gives
p²-4q<0,
4q > p²
q > 1/4 p².
Hope that helps.

Answer 4

1) Since the cross term is -4x, you know that p=2, and so
x^2 -4x +8 = (x-2)^2 +4
The roots of the equation occur at
x=(1/2)*(4 +/- sqrt(16-4(8)(1)))
x=(1/2)*(4 +/- sqrt(-16))
Therefore the roots are complex, f(x) is never equal to zero in the real plane, and since f(x) is an upward quadratic, all values must be positive for all x.

Answer 5

1) f(x)=x²-4x+8=(x²-2*2x+2²)-2²+8
f(x)=(x-2)²+4

2) D=b²-4ac<0 and a>0 so
9-16a<0 and a>0
16a>9 and a>0
a>9/16 and a>0 so
a E (9/16,+oo)

3)a=-1<0 and D=p²-4q
expression -x²+px-q is negative for all real values of x if D<0
so p²-4q<0 <=> q > 1/4 p²
_______________________________________
expression ax²+bx+c for all real values of x is:
i) negative if a ii) positive if a>o and D=b²-4ac<0

Answer 6

1. factor it out... (x-2)^2 + 4

2. For any number where a > 0

3. This will work - plug in values of different amounts, and try it.