If p and q are two prime numbers whose sum of the squares is an odd integer (p

Question

If p and q are two prime numbers whose sum of the squares is an odd integer (p<q) and the roots of the quadratic (2 p +1) x 2+( p + q -7) x -5=0 are equal in magnitude but opposite in sign then | p - q | is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Since p and q are prime number and p 2+ q 2= odd integer and also p < q Hence, p=2 5 x2+(q-5) x-5=0 text S.O.R. =0 (-(q-5)/5)=0 ⇒ q=5 ∴ |p-q|=3

Q. If $p$ and $q$ are two prime numbers whose sum of the squares is an odd integer $(p < q)$ and the roots of the quadratic $(2 p + 1) x^{2} + (p + q - 7) x - 5 = 0$ are equal in magnitude but opposite in sign then $∣ p - q ∣$ is equal to

1

2

3

4

Since $p$ and $q$ are prime number and $p^{2} + q^{2} =$ odd integer and also $p < q$
Hence, $p = 2$
$5 x^{2} + (q - 5) x - 5 = 0$
$S.O.R. = 0$
$\frac{- ( q - 5 )}{5} = 0 \Rightarrow q = 5$
$∴ ∣ p - q ∣ = 3$

Q. If p and q are two prime numbers whose sum of the squares is an odd integer (p<q) and the roots of the quadratic (2p+1)x2+(p+q−7)x−5=0 are equal in magnitude but opposite in sign then ∣p−q∣ is equal to

1

2

3

4

Since p and q are prime number and p2+q2= odd integer and also p<q Hence, p=2 5x2+(q−5)x−5=0 S.O.R. =0 5−(q−5)​=0⇒q=5 ∴∣p−q∣=3

Q. If $p$ and $q$ are two prime numbers whose sum of the squares is an odd integer $(p < q)$ and the roots of the quadratic $(2 p + 1) x^{2} + (p + q - 7) x - 5 = 0$ are equal in magnitude but opposite in sign then $∣ p - q ∣$ is equal to

Since $p$ and $q$ are prime number and $p^{2} + q^{2} =$ odd integer and also $p < q$
Hence, $p = 2$
$5 x^{2} + (q - 5) x - 5 = 0$
$S.O.R. = 0$
$\frac{- ( q - 5 )}{5} = 0 \Rightarrow q = 5$
$∴ ∣ p - q ∣ = 3$