Question

If x = √p+q + √p-q/√p+q- √p-q,then find the value of qx^2-2px+q

Answer 1

x= √p + q + √p - q /  √p + q - √p - q

rationalizing
 x =  2p + √p² - q²  /  2q =  p + √p² - q²  /  q
⇒ xq - p  =  √p² - q² 

squaring both sides

⇒  q²x² + p² - 2qxp =  p² - q² ⇒  q²x²  -2pqx +q²  = 0
⇒  q( qx²-2px + q ) = 0 

therefore ans = 0

hope u'll get ...........if any doubt , ask

Answer 2

$q \times x^{2} - 2 \times p \times x + q = 0$

Given:

$\mathrm { x } = \frac { \sqrt { \mathrm { p } + \mathrm { q } } + \sqrt { \mathrm { p } - \mathrm { q } } } { \sqrt { \mathrm { p } + \mathrm { q } } - \sqrt { \mathrm { p } - \mathrm { q } } }$

To find:

Calculate the value of $qx^2 -2px+q$

Answer:

$\mathrm { x } = \frac { \sqrt { \mathrm { p } + \mathrm { q } } + \sqrt { \mathrm { p } - \mathrm { q } } } { \sqrt { \mathrm { p } + \mathrm { q } } - \sqrt { \mathrm { p } - \mathrm { q } } }$

Rationalize the above equation

$x = \frac { \sqrt { p + q } + \sqrt { p - q } } { \sqrt { p + q } - \sqrt { p - q } } \times \frac { \sqrt { p + q } + \sqrt { p - q } } { \sqrt { p + q } + \sqrt { p - q } }$

$x = \frac { ( \sqrt { p + q } + \sqrt { p - q } ) ^ { 2 } } { ( \sqrt { p + q } ) ^ { 2 } - ( \sqrt { p - q } ) ^ { 2 } }$

$x = \frac { ( \sqrt { p + q } ) ^ { 2 } + ( \sqrt { p - q } ) ^ { 2 } + 2 \sqrt { p + q } \sqrt { p - q } } { p + q - p + q }$

$x = \frac {p + q + p – q + 2 \times \sqrt {{p}^{2} - {q}^{2}}} {2 \times q}$

$x = \frac { 2 \times p + 2 \times \sqrt {{p}^{2} - {q}^{2}}} {2 \times q}$

$x = \frac { p + \sqrt {{p}^{2} - {q}^{2}}} {q}$

$q \times x - p = \sqrt {{p}^{2} - {q}^{2}}$

${q \times x - p}^{2} = p^{2} -q^{2}$

${q \times x}^2 + p^{2} - 2 \times p \times q \times x - p^{2} + q^{2} =0$

${q \times x}^2 - 2 \times p \times q \times x  + q^{2} =0$

${q \times x^{2} - 2 \times p \times x +q}} =0$

Obviously,

x= root of quadratic:

Thus,

$q \times x^{2} - 2 \times p \times x + q = 0$