Question

hey


PLZZ I NEED IT NOWWW....



if the roots of the equation
$x^{2} +px +q =0$

differ from the roots pf the equation
$x^{2} + qx +p = 0$

by the same quantity
then

p+q =?

ANSWER IS ???
NEED FULL EXPLANATION

Answer 1

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

Therefore, p + q = -4

Answer 2

Step-by-step explanation:

Firslty we have given that

${x}^{2} + px + q = 0$

and also

${x}^{2} + qx + p = 0$

than From Eqn. 1

$q = - {x}^{2} - px$

and from Eq.2

$p = - {x}^{2} - qx$

Now

we have to Find

$p + q = - {x}^{2} - px - {x}^{2} - qx$

$= - 2 {x}^{2} - px - qx$

$= x( - 2x - p - q)$

So This Is the Value of P+q ...

Thanks.