Question

lcm 300and hcf 15 and other numbers is less than 50 find two numbers​

Answer 1

let the numbers be x and x-50 ,

then APQ,

LCM =300, HCF=15

now we know that

.

so...

300*15=x * (x-50)

${x}^{2} - 50x - 4500 = 0$

now we check if the equation has real values of x or not.

so here a= 1, b= -50 c= -4500

and for real roots to exist..

$b^{2} - 4ac \ \geqslant 0$

so here..

2500-4*1*(-4500)

20500>0

so the equation has real roots..

now there are two real roots for x

and they are

..

x= [-b+√(b^2-4ac)]/2a

and

x= [-b-√(b^2-4ac)]/2a

on solving we get x = -5( -5 + √205)

and 5(5+√205) .

now you also know that this is supposed answer for your question but I think you messed up somewhere in giving the numbers..

cheers