Question

abx²+(b²-ac)x-bc =0. Find the value of X.

Answer 1

Here, a=ab

b= b2-4ac

c= -bc

Now, b2-4ac

= (b2-ac)2-4.ab.(-bc)

= b4+a2c2-2b2ac+4b2ac

= b4+a2c2+2b2ac

= (b2+ac)2

Now, √b2-4ac= b2+ac

Putting it in quadratic formula,

x=(-b ± √b2-4ac)/2a

=> x=(-b2+ac±b2+ac)/2ab

=> x=-b2+ac+b2+ac/2ab or -b2+ac-b2+ac/2ab

=> x= 2ac/2ab or -2b2+2ac/2ab

=> x= c/b or x= 2(-b2+ac)/2ab or x= ac-b2/ab

Hope it helps!

Answer 2

Hii friend,

This given equation is abX²+(b²-ac)X-bc = 0

This equation is in the form of Ax²+BX+C

THEREFORE,

A = ab , B = (b²-ac) and C = -bc

THEREFORE,

D = ✓B²-4AC = (b²-ac)² + 4ab² c

= b⁴+a²c²-2ab²c + 4ab²c

= b⁴+ a²c² + 2ab²c = (b²+ac) > 0

So , the given equation has real roots.

Now,

✓D = (b²+ac)

THEREFORE,

Alpha = -B+✓D/2A = -(b²-ac) + (b²+ac) /2ab = 2ac/2ab = c/b , and

Beta = -B-✓D/2A = -(b²-ac) + (b²+ac)/2ab = -2b²/2ab = -b/a


HENCE,

c/b and -b/a are the roots of the given equation.


HOPE IT WILL HELP YOU..... :-)