Question

If the roots of the equation
( c2- ab) x2 +2(a2-bc) x + b²-ac=0
are equal, prove that either a = 0 or a3+c3= 3abc​

Answer 1

Answer:

Compare given Quadratic equation (c²-ab)x²+2(a²-bc)x+b²-ac=0 with Ax²+Bx+C =0 ,we get

A = (c²-ab),

B = 2(a²-bc),

C = b²-ac,

Discreminant (D) = 0

/* Equal roots *

=> B²-4AC = 0

=> [-2(a²-bc)]²-4(c²-ab)(b²-ac)=0

=> 4(a²-bc)²-4(b²c²-ac³-ab³+a²bc)=0

=> (a²-bc)²-(b²c²-ac³-ab³+a²bc)=0

=> a⁴-2a²bc+b²c²-b²c²+ac²+ab³-a²bc=0

=> a⁴-3a²bc+ac³+ab³=0

=> a(a³-3abc+c³+b³)=0

=> a=0 Or (a³+b³+c³-3abc) = 0

=> a=0 Or a³+b³+c³=3abc

•••♪

Answer 2

ANSWER:

Given:

•Equation is (c² -ab)x² - 2(a²- bc)x+(b²-ac)=0

To prove:

•a= 0 or

•a³ + b³ +c³= 3abc

Proof:

We have,

⚫A= (c² -ab)

⚫B= -2(a² -bc)

⚫C= (b² -ac)

It is given that the equation has real & equal roots.

Therefore,

D=0

We know the formula of the quadratic equation:

=) b² - 4ac=0

Putting the value of a, b,& c in above equation, we get;

=) [-2(a²-bc)]² -4(c² -ab)(b²-ac)=0

=) 4(a² -bc)² -4(c²b² -ac³ -ab³ +a²bc)=0

=)4(a⁴ +b²c² -2a²bc) -4(c²b² -ac³-ab³+a²bc)=0

=)a⁴+b²c²-2a²bc-b²c² +ac³ +ab³ -a²bc=0

=) a⁴ +ab³ +ac³ -3a²bc =0

=) a(a³ +b³ +c³ -3abc) =0

=) a=0 or a³ +b³ +c³ -3abc. [proved]

Thank you.