Question

If a(b-c) x^2 + b(c-a) x + c(a-b) = 0 has equal roots, show that 2/b = 1/a + 1/c​

Answer 1

Answer:

→ Delta=0

→ [b(c-a)]^2–4* ac(a-b)(b-c)=0

→ [b(c-a)]^2=4* ac(a-b)(b-c)

→ b^2(c-a)^2=4* ac(a-b)(b-c)

→ b^2(c^2–2ac+a^2)=4* ac(ab-b^2-ac+bc)

→ b^2c^2–2ac b^2+a^2b^2=4* ac(ab-b^2-ac+bc)

→ b^2c^2+2ac b^2+a^2b^2=4* ac(ab-ac+bc)

→ b^2(c+a)^2=4* ac(ab-ac+bc)

→ [b(c+a)]^2=-4* (ac)^2 + 4ac*b(c+a)

→ [b(c+a)]^2- 4ac*b(c+a)+4* (ac)^2 = 0

→ ([b(c+a)]- 2ac)^2 = 0

→ [b(c+a)]- 2ac = 0

→ b(c+a)= 2ac

Given quadratic equation,

a(b-c)x^2+b(c-a)^2+c(a-b)=0

also zeroes of given equation are equal.

Therefore,

discriminant=0

b^2-4ac=0

{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0

b^2(c^2+a^2-2ac)-4{ab-ac}{ac-bc}=0

b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0

b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0

We know that,

a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2

By above identity we get,

{bc+ab-2ac}^2=0

bc+ab-2ac=0

b(a+c)=2ac

b=2ac/(a+c)

Hence if zeroes of given quadratic equations then,

b=2ac/(a+c)

Answer 2

$\sf{ \bold{ Given \: Quadratic \: Equation \: - }} \\ \\ \sf{ \implies{ a(b-c) x^2 + b(c-a) x + c(a-b) = 0 }} \\ \\ \sf{ \implies {( ab - ac) x^2 + ( bc - ac ) x + ( ac - bc ) = 0 }} \\ \\ \sf{ Now , \: we \: have \: the \: following \: condition \: - }$

$\\ \\ \sf{ This \: Quadratic \: Equation \: has \: equal \: roots . } \\ \\ \sf{ \bold{ Hence \: - }} \\ \\ \sf{ The \: value \: of \: the \: discriminant \: is \: zero } \\ \\ \sf{ We \: know \: that \: - } \\ \\ \sf{ D = b^2 - 4ac } \\ \\ \sf{ Substituting \: the \: given \: values \: - } \\ \\ \sf{ D = {( bc - ac )} ^ 2 - 4 ( ab - ac )( ac - bc ) }$

$\\ \\ \sf{ D = b^2c^2 + a^2 c^2 - 2abc^2 - 4 { ( a^2 bc - ab^2c - a^2 c^2 + abc^2 ) } } \\ \\ \sf{ => D = b^2c^2 + a^2 c^2 - 2abc^2 - 4a^2bc - 4ab^2c - 4a^2c^2 + 4abc^2 } \\ \\ \sf{ => D = {(bc + ab - 2ac) } ^ 2 } \\ \\ \sf{ But \: D = 0 } \\ \\ \sf{ \bold{ Hence \: - }} \\ \\ \sf{ \implies { {(bc + ab - 2ac) } ^ 2 = 0 }} \\ \\ \sf{ \implies{ (bc + ab - 2ac) = 0 }} \\ \\ \sf{ \implies {b ( a + c ) = 2 ac }} \\ \\ \sf{ \implies { \dfrac{ ( a + c ) }{ ac } = \dfrac{ 2 }{ b } }} \\ \\ \sf{ \implies { \dfrac{ 2 }{ b } = \dfrac{ 1 }{ a } + \dfrac{ 1 }{ c } }} \\ \\ \sf{ \bold{ Hence \: Proved \: !!!! }}$