If the roots of the equation (c²-ab)x²- 2(a²-bc) x +(b²-ac) =0 are real and equal show that either a = 0 or (a³+b³ +c³ ) =3ABC.
Answers
Answer:
a = 0 Or a³ + b³ + c³ = 3abc
Step-by-step-explanation:
The given quadratic equation is
( c² - ab ) x² - 2 ( a² - bc ) x + ( b² - ac ) = 0.
We have given that,
The roots of the equation are real and equal.
We have to show that either a = 0 or ( a³ + b³ + c³ ) = 3abc.
Now,
( c² - ab ) x² - 2 ( a² - bc ) x + ( b² - ac ) = 0
Comparing with px² + qx + r = 0, we get,
- p = ( c² - ab )
- q = - 2 ( a² - bc )
- r = ( b² - ac )
We know that,
For real and equal roots,
q² - 4pr = 0
⇒ [ - 2 ( a² - bc ) ]² - [ 4 * ( c² - ab ) * ( b² - ac ) ] = 0
⇒ ( - 2 )² ( a² - bc )² - [ ( 4c² - 4ab ) ( b² - ac ) ] = 0
⇒ 4 [ ( a² )² - 2a²bc + ( bc )² ] - [ 4c² ( b² - ac ) - 4ab ( b² - ac ) ] = 0
⇒ 4 ( a⁴ - 2a²bc + b²c² ) - ( 4b²c² - 4ac³ - 4ab³ + 4a²bc ) = 0
⇒ ( 4a⁴- 8a²bc + 4b²c² ) - ( 4b²c² - 4ac³ - 4ab³ + 4a²bc ) = 0
⇒ 4a⁴ - 8a²bc + 4b²c² - 4b²c² + 4ac³ + 4ab³ - 4a²bc = 0
⇒ 4a⁴ - 8a²bc - 4a²bc + 0 + 4ac³ + 4ab³ = 0
⇒ 4a⁴ - 12a²bc + 4ac³ + 4ab³ = 0
⇒ 4a ( a³ - 3abc + c³ + b³ ) = 0
⇒ 4a = 0 OR ( a³ - 3abc + c³ + b³ ) = 0
⇒ a = 0 ÷ 4 OR a³ - 3abc + c³ + b³ = 0
⇒ a = 0 OR a³ + b³ + c³ = 3abc
∴ Either a = 0 Or a³ + b³ + c³ = 3abc
Hence shown!
Answer:
Step-by-step explanation:
