Question

$\tt \: find \:number \: of \: orered \: pairs \:( a \: b) \: such \: that \: the \: equation \: x^{2} + ax + b = 0. \: has \: rational \: roots. \\ given \: \: a \: b \: are \: the \: odd \: natural \: number.$
easy and conceptual question. solve it​

Answer 1

We're given,

$x^2+ax+b=0$ has rational roots.

The roots of ${x}^{2} + ax + b = 0$ are, $\dfrac{-a\pm\sqrt{a^2-4b}}{2}.$

In order for the roots to be rational, the discriminant, $D=a^2-4b$ shall be a perfect square.

So, for a natural number $k,$

$a^2-4b=k^2$

It is given that, a and b are odd natural numbers.

• Then, $a^2$ shall also be odd.
• And $4b$ shall be even.
• We know that, odd - even (in LHS) = odd (in RHS). Hence $k^2$ is odd.
• From here we conclude, $k$ is an odd natural number.

$\longrightarrow a^2-k^2=4b$

$\longrightarrow (a + k)(a - k) = 4b$

Case 1:

If $(a-k)=b,$

• We know that, $a,b$ and $k$ are odd natural numbers.
• We know that, odd - odd (in LHS) = even (in RHS)
• But $b$ is odd.
• That's a contradiction.

Hence no ordered pairs of (a, b) satisfying case 1.

Case 2:

If $(a+k)=b,$

• We know that, $a,b$ and $k$ are odd natural numbers.
• We know that, odd + odd (in LHS) = even (in RHS)
• But $b$ is odd.
• That's a contradiction.

Hence no ordered pairs of (a, b) satisfying case 2.

Hence, there are no ordered pairs satisfying this condition.