Q 9

In the proof of Theorem $12.45$ we mentioned that if the quantity $A C - B^{2} > 0$ , then the signs of $A$ and $C$ are the same. Explain why.

Verified

Answer

If $A$ and $C$ have opposite signs then both terms $A C$ and $- B^{2}$ are negative, so their sum is negative.

So that $A C - B^{2} > 0$ , then the signs of $A$ and $C$ are same.

1Step 1. Given Information

We have given that, In a theorem we stated that :-

If $A C - B^{2} > 0$ , then $A$ and $C$ are of same sign.

Here we have explain the reason behind this statement.

2Step 2. Explanation of required reason

Let $A$ and $C$ are of opposite sign.

We know that multiplication of two opposite sign numbers is always negative.

That is $A C$ is negative.

Also we know that :-

$B^{2}$ is always positive.

That is $- B^{2}$ will be negative.

The sum of two negatives is again negative.

That is :-

$A C - B^{2} < 0$ .

In short we find that if $A$ and $C$ are of opposite signs. Then $A C - B^{2} < 0$ .

So this quantity is positive if and only if $A a n d C$ are of same sign.
Hence proved.