Question

((x^(a+b) )^2 (x^(b+c) )^2 (x^(c+a) )^2)/(x^a x^b x^c )^4 =1 PROVE THIS QUESTION FAST!!!!! Correct answered would be marked as brainliest !!!!!!

Answer 1

Answer:

Given: (xa/xb)^1/ab( xb /xc)^1/bc(xc/xa)1/ca

We need to prove the gives equation

LHS=(xa/xb)^1/ab( xb /xc)^1/bc(xc/xa)1/ca

Using laws of exponents

= (xa/xb)1/ab( xb /xc)1/bc(xc/xa)1/ca

= x(a-b)/ab * x^(b-c)/bc * x^(c-a)/ca

= x[(a-b)/ab + (b-c)/bc + (c-a)/ca]

= x[c(a-b)/abc + a(b-c)/abc + b(c-a)/abc ]

= x { [c(a-b)+ a(b-c) + b(c-a) ]/abc }

= x ( ac – bc + ab – ac + bc – ab ] /abc

= x 0/abc

= x0

= 1

= RHS

Hence proved

Answer 2

Answer:

We have,

(x−a)(x−c)+2(x−b)(x−d)=0

⇒3x

2

−(a+c+2b+2d)x+(ac+2bd)=0

Discriminant

={(a+2d)+(c+2b)}

2

−4.3(ac+2bd)

={(a+2d)+(c+2b)}

2

−12(ac+2bd)

={(a+2d)−(c+2b)}

2

−4(a+2d)(c+2b)−12(ac+2bd)

={(a+2d)−(c+2b)}

2

+4ac+8ab+8cd+16bd−12ac−24bd

={(a+2d)−(c+2b)}

2

+8(ab+cd−ac−bd)

={(a+2d)−(c+2b)}2+8(c−b)(d−a)>0

∵a<b<c<d⇒c−b>0and d−a>0

∴ The roots of the given equation are real and distinct