Question

pls pls answer this....
☺❤​

Answer 1

as the roots of above eq. are real and unequal, therefore

D>0

${1}^{2} - 4p.q > 0 \\ = pq < \frac{1}{4}$

now in the equation

${x}^{2} - 4 \sqrt{p.q} x + 1 = 0 \\ d = {(4.( \sqrt{pq}))}^{2} - 4(1).1 \\ = 16pq - 4 \\ now \: for \: the \: eq. \: to \: have \: complex \: roots \\ d < 0 \\ = pq < \frac{1}{4} \\ which \: is \: true \: as \: we \: have \: proved \: it \: earlier$

Answer 2

as the roots of above eq. are real and unequal, therefore

D>0