Question

if the roots of the equation x2+px+q=0 differ from the roots of tue equation x2+qx+p=0 by the same quantity then

A. P+Q+1
B. P+Q+2
C. P+Q+4

Answer 1

well thanks for providing options

P(x) = x² + px +q = 0
let's its root be a and b

then sum of roots = a + b = -p
       product of roots = ab = q

difference of roots = a-b = √(a + b)² - 4ab = √(-p)² - 4q = √p² - 4q ..........(1)


now P(x) = x² + qx + p
let's its roots be c and d

sum of roots = c + d = -q
products of roots = p 
difference of roots = c-d = √(c+d)² - 4 cd = √(-q)² - 4p = √q² - 4p ......(2)

now A/Q  difference of roots of both equations are equal

therefore eq (1) = eq (2)

⇒√p² - 4q = √q² - p 
squaring both sides

⇒p² -4q = q² - 4p ⇒ p² - q² = 4(q - p)
⇒ (p-q)(p+q) - 4(q - p) = 0
⇒ (p - q) (p + q +4 ) = 0

⇒ p=q     or  p + q + 4=0

since both equation are different therefore p≠q
and p + q + 4 = 0

hence option c) p + q + 4 = 0

hope u'll get it

if any queries, then ask

Answer 2

Answer:(c)

Step-by-step explanation:

P(x) = x² + px +q = 0

let's its root be a and b

then sum of roots = a + b = -p

       product of roots = ab = q

difference of roots = a-b = √(a + b)² - 4ab = √(-p)² - 4q = √p² - 4q ..........(1)

now P(x) = x² + qx + p

let's its roots be c and d

sum of roots = c + d = -q

products of roots = p 

difference of roots = c-d = √(c+d)² - 4 cd = √(-q)² - 4p = √q² - 4p ......(2)

now A/Q  difference of roots of both equations are equal

therefore eq (1) = eq (2)

⇒√p² - 4q = √q² - p 

squaring both sides

⇒p² -4q = q² - 4p ⇒ p² - q² = 4(q - p)

⇒ (p-q)(p+q) - 4(q - p) = 0

⇒ (p - q) (p + q +4 ) = 0

⇒ p=q     or  p + q + 4=0

since both equation are different therefore p≠q

and p + q + 4 = 0

hence option c) p + q + 4 = 0