Question

The quadratic $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be equal to

• A. $50$
• B. $37$
• C. $61$
• D. $19$

Answer 1

Answer: (A)

$50$

Let $\alpha$ and $\beta$ be the roots of the given quadratic equation.
Now $\alpha \beta =b+1$.....(1)
And $\alpha +\beta =-a$......(2)

Now $a^2+b^2={(\alpha +\beta )}^2+{(\alpha \beta -1)}^2$
The above equation can be written as:
$a^2+b^2=({\alpha }^2+1)({\beta }^2+1)$ with the help of (1) and (2)

Now $({\alpha }^2+1)({\beta }^2+1)$ is clearly product of two natural numbers (each of which is not equal to $1$).
Hence, the answer can't be a prime number.

$a^2+b^2$ can be split into factors

$37,61,19$ cannot be split into factors as they are primes.

Therefore, $a^2+b^2$ can be equal to $50$.