Question

PLS help me guys...............

​hint:use quadratic formula for this

Answer 1

Step-by-step explanation:

${p}^{2} {x}^{2} + ( {p}^{2} - {q}^{2} )x - {q}^{2} = 0$

Compare the equation with standard Quadratic equation

$a {x}^{2} + bx + c = 0 \\ \\ here \\ \\ a = {p}^{2} \\ \\ b = ( {p}^{2} - {q}^{2} ) \\ \\ c = - {q}^{2} \\ \\ quadratic \: formula x_{1,2}= \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\x_{1,2} = \frac{ (- {p}^{2} + {q}^{2} ) ± \sqrt{ ({{p}^{2} - {q}^{2}})^{2} + 4 {p}^{2} {q}^{2} } }{2 {p}^{2} } \\ \\ x_{1,2}= \frac{ - {p}^{2} + {q}^{2} ± \sqrt{ {p}^{4} + {q}^{4} - 2 {p}^{2} {q}^{2} + 4 {p}^{2} {q}^{2}} }{2 {p}^{2} } \\ \\x_{1,2}=\frac{ - {p}^{2} + {q}^{2} ± \sqrt{ {p}^{4} + {q}^{4} + 2 {p}^{2} {q}^{2}} }{2 {p}^{2} } \\ \\ x_{1,2}= \frac{ - {p}^{2} + {q}^{2} ± \sqrt{ {( {p}^{2} + {q}^{2} )}^{2} }}{2 {p}^{2} } \\ \\ x_{1,2} = \frac{- {p}^{2} + {q}^{2} ± {p}^{2} + {q}^{2}}{2 {q}^{2} } \\ \\ x_1 = \frac{2 {q}^{2} }{ 2{q}^{2} } \\ \\ x_1= 1 \\ \\ x_2 = - \frac{ {p}^{2} }{ {q}^{2} }$

Hope it helps you

Answer 2

Answer:

ok

okStep-by-step explanation:

SOLUTION

SOLUTIONBRAINLIEST BUDDY