Quadratic equations problems

May 2010
8
0
Hello,
Can anyone please help me through these problems? Even a hint will suffice. I need it at the earliest. Thanks.


1. 1 is a root of the equation ax2+bx+c, where a,b,c belong to R. Prove that the roots of 4ax2+3bx+2c are real and unequal.

2. If x2+ax+b belongs to Z for all x->Z and roots are rational. Prove that the roots are infact integers.


Please do help. Thanks again,
Sagar
 
Nov 2009
927
260
Wellington
Hello,
for the number (1), consider the discriminant of the quadratic equation : if it is strictly positive, then the quadratic will have two distinct real roots.

For the number (2), work around the discriminant : if it is a perfect square, the equation will have rational roots, and if the leading coefficient of \(\displaystyle x^2\) is equal to 1, then the roots will actually be integers. Prove this by using the quadratic formula and showing that the upper part of the fraction is always even (therefore dividing it by two yields an integer).

Does it make sense ?
 
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May 2010
8
0
Thanks for the hints.

(1), the discriminant turns out to be 9b2-32ac, but I don't know how to proceed further and prove that it is positive. ( The first equation gives a+b+c=0, if 1 is the root, but I do not know how to use it)

(2), could you tell me how to prove the numerator of the quadratic formula will be even always?

Thanks.
 
Nov 2009
927
260
Wellington
1 is a root of the equation ax2+bx+c, where a,b,c belong to R. Prove that the roots of 4ax2+3bx+2c are real and unequal.
Since 1 is a root of \(\displaystyle ax^2 + bx + c\), \(\displaystyle x - 1\) must be a factor of this quadratic. Note that :

\(\displaystyle (x - 1)(ax - c) = ax^2 - x(c + a) + c\)

This in turn implies that \(\displaystyle b = -c - a\) (which is, notice, equivalent to \(\displaystyle a + b + c = 0\) ;)). Now taking the discriminant of the second equation :

\(\displaystyle \Delta = 9b^2 - 32ac\)

You can substitute the result we found before :

\(\displaystyle \Delta = 9(- c - a)^2 - 32ac\)

\(\displaystyle \Delta = 9(c^2 + 2ac + a^2) - 32ac\)

\(\displaystyle \Delta = 9c^2 + 18ac + 9a^2 - 32ac\)

\(\displaystyle \Delta = 9a^2 + 9c^2 - 14ac\)

\(\displaystyle \Delta = 9(a^2 + c^2) - 14ac\)

Here a little bit of analysis is required : if \(\displaystyle a\) and \(\displaystyle c\) have different signs, then the discriminant must be positive. What if they do have same signs ? Well this is the neat trick : you only have to prove that for any \(\displaystyle a\), \(\displaystyle c\), the following does not hold (proof by contradiction) :

\(\displaystyle 9(a^2 + c^2) < 14ac\)

That is :

\(\displaystyle a^2 + c^2 < \frac{14}{9} ac\)

You may want to complete the square, that is, take exactly \(\displaystyle 2ac\) from the right side :

\(\displaystyle a^2 + c^2 - 2ac < \frac{14}{9} ac - 2ac\)

\(\displaystyle a^2 - 2ac + c^2 < \frac{14}{9} ac - 2ac\)

Then we have :

\(\displaystyle (a - c)^2 < \frac{14}{9} ac - 2ac\)

Factorizing on the other side :

\(\displaystyle (a - c)^2 < \left (\frac{14}{9} - 2 \right ) ac\)

\(\displaystyle (a - c)^2 < \frac{-4}{9} ac\)

But ! Since we assumed \(\displaystyle a\) and \(\displaystyle c\) are of same signs, \(\displaystyle ac\) must be positive. Therefore, \(\displaystyle \frac{-4}{9} ac\) must be negative. However, look at the left-hand-side : a square can only be positive in \(\displaystyle \mathbb{R}\) ! We reached a contradiction which invalidates the original statement which was \(\displaystyle 9(a^2 + c^2) < 14ac\). This implies that \(\displaystyle 9(a^2 + c^2) > 14ac\), which in turn implies that \(\displaystyle 9(a^2 + c^2) - 14ac > 0\), therefore the discriminant is always positive.

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If x2+ax+b belongs to Z for all x->Z and roots are rational. Prove that the roots are infact integers.
Let \(\displaystyle x^2 + ax + b\) be a quadratic equation with rational roots. This means that \(\displaystyle a^2 - 4b = k^2\) for some integer \(\displaystyle k\).

Now assume \(\displaystyle a\) and \(\displaystyle b\) are both even.

\(\displaystyle a^2\) will be even, \(\displaystyle 4b\) will (obviously) be even, so \(\displaystyle a^2 - 4b\) will be even. This implies that \(\displaystyle \sqrt{a^2 - 4b}\) be even as well (since \(\displaystyle k\) is an integer and \(\displaystyle k^2\) keeps the parity of \(\displaystyle k\)). Looking at the quadratic formula :

\(\displaystyle \frac{-a \pm \sqrt{a^2 - 4b}}{2}\)

We know that \(\displaystyle -a\) will be even, and so is the square root as shown precedently, therefore \(\displaystyle -a \pm \sqrt{a^2 - 4b}\) will be even. This implies that \(\displaystyle \frac{-a \pm \sqrt{a^2 - 4b}}{2}\) will be an integer.

There are three cases left : (\(\displaystyle a\) even, \(\displaystyle b\) odd), (\(\displaystyle a\) odd, \(\displaystyle b\) even), (\(\displaystyle a\) odd, \(\displaystyle b\) odd). I'll let them to you, if you still need help don't hesitate.

(Whew)
 
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