Question

Previous year Question of jee mains

mathamatics shift :1
24 feb 2021​

Answer 1

Answer:

Step-by-step explanation:

$(p+q)^2 =p^2+q^2+2pq\\\\(2)^2 =p^2+q^2+2qp\\\\4-2pq =p^2+q^2\\\\taking \;square \; on \; both \;sides\\(4-2pq)^2=(p^2+q^2)^2\\\\16+4p^2q^2 -16pq= p^4+q^4+2p^2q^2\\\\16+4p^2q^2 -16pq = 272+2p^2q^2\\\\2p^2q^2-16pq-256=0\\\\solve \;this\; quadretic \;for \; pq\\you \; will \; get \\pq=16\\\\so \;answer\; is \;3$

Answer 2

Answer:

p+q = 2 ------(1)

squaring both sides,,

(p+q)² = 2²

→ p²+q²+2pq = 4

→ p²+q² = 4-2pq --------(2)

p⁴+q⁴ = 272

$→ ({ {p}^{2} })^{2} + ({ {q}^{2} })^{2} = 272$

$→ {( {p}^{2} + {q}^{2} ) }^{2} - 2 {p}^{2} {q}^{2} = 272$

from eqn (2) , we get,

$→ {(4 - 2pq)}^{2} - 2 {p}^{2} {q}^{2} = 272$

$→16 + 4 {p}^{2} {q}^{2} - 16pq - 2 {p}^{2} {q}^{2} = 272$

$→2 {p}^{2} {q}^{2} - 16pq + 16 - 272 = 0$

$→2 {p}^{2} {q}^{2} - 16pq - 256 = 0$

$→ {p}^{2} {q}^{2} - 8pq - 128 = 0$

let pq = z

$→ {z}^{2} - 8z - 128 = 0$

$→ {z}^{2} - 16z + 8z - 128 = 0$

$→z(z - 16) + 8(z - 16) = 0$

$→(z - 16)(z + 8) = 0$

$→z = 16 \: or \: - 8$

$→pq \: = 16 \: or \: - 8$

since, p and q are two positive number so 'pq' cannot be negative.

thus, pq = 16 -----(3)

So, the quadratic equation whose roots are p and q is :

x²-(p+q)x+pq = 0

from eqn (1) and (3) :

→ x²-2x+16 = 0 is the req quadratic eqn.