Solve for x: (i) `10x^(2)-27x+5=0` (ii) `pqx^(2)-(p^(2)+q^(2))x+pq=0`

Answer

(i) `(5)/(2), (1)/(5)`
(ii) `(p)/(q), (q)/(p)`

Explanation

(i) `10 x^(2) - 27x +5=0` ltBrgt `x= (-bpm sqrt(b^(2) - 4ac))/(2a) = (27 pm sqrt((-27)^(2) - 4(10)(5)))/(20)`
`= (27pm 23)/(20 ) rArr x =(5)/(2), (1)/(5)`
(ii) `(pq)x^(2) - (p^(2) + q^(2))x+(pq)=0`
`x= ((p^(2) + q^(2)) pm sqrt((p^(2)+ q^(2))^(2) - 4(pq)(pq)))/( 2pq)`
`= ((p^(2) + q^(2)) pm (p^(2) - q^(2)))/( 2pq) rArr x = (p)/(q), (q)/(p)`