Question

"If the roots of the equation
a(b-c) x^2+b(c-a) x+c(a-b) =0Are equal then show that 2/b=1/a+1/c​

Answer 1

GIVEN :–

• A quadratic equation a(b-c) x² + b(c-a)x + c(a-b) =0 have equal roots.

TO PROVE :–

$\\ \sf \implies \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

SOLUTION :–

• If roots of a quadratic equation are equal then Discriminant –

$\\ \longrightarrow \large { \boxed{ \sf D = 0}}$

$\\ \longrightarrow \large { \boxed{ \sf {b}^{2} - 4ac = 0}}$

• Here –

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: \sf \: a = a(b - c)$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: \sf \: b = b(c-a)$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: \sf \: c = c(a-b)$

• So that –

$\\ \implies \sf {b}^{2} - 4ac = 0$

$\\ \implies \sf {b}^{2} = 4ac$

• Now put the values –

$\\ \implies \sf { \{b(c - a) \}}^{2} = 4 \{a(b - c) \} \{c(a - b) \}$

$\\ \implies \sf { {b}^{2} (c - a) }^{2} = 4ac(b - c)(a - b)$

$\\ \implies \sf {b}^{2} ( {c}^{2} + {a}^{2} - 2ac ) = 4ac(ba - {b}^{2} - ca + bc)$

$\\ \implies \sf {b}^{2} {c}^{2} + {b}^{2} {a}^{2} - 2a {b}^{2}c= 4 {a}^{2}bc - 4a {b}^{2}c - 4 {a}^{2} {c}^{2} +4ab {c}^{2}$

$\\ \implies \sf {b}^{2} {c}^{2} + {b}^{2} {a}^{2} + 2a {b}^{2}c - 4 {a}^{2}bc + 4 {a}^{2} {c}^{2} - 4ab {c}^{2} = 0$

• We should write this as –

$\\ \implies \sf {(bc + ba)}^{2} - 4ac(ab - ac + bc)= 0$

• Now –

$\\ \implies \sf {(bc + ba)}^{2} - 4ac(ab - ac + bc + ac - ac)= 0$

$\\ \implies \sf {(bc + ba)}^{2} - 4ac(ab - 2ac + bc ) - 4 {(ac)}^{2} = 0$

$\\ \implies \sf {(bc + ba)}^{2} - {(2ac)}^{2} = 4ac(ab - 2ac + bc )$

• Using identity –

$\\ \longrightarrow \: \: \sf {a}^{2} - {b}^{2} = (a + b)(a - b)$

• So that –

$\\ \implies \sf { \cancel{(bc + ba - 2ac)}}(bc + ba + 2ac)= 4ac{ \cancel{(ab - 2ac + bc )}}$

$\\ \implies \sf (bc + ba + 2ac)= 4ac$

$\\ \implies \sf bc + ba = 2ac$

• Now divided by (abc) on both sides –

$\\ \implies \sf \dfrac{bc + ba}{abc} = \dfrac{2ac}{abc}$

$\\ \implies \sf \dfrac{bc}{abc} + \dfrac{ba}{abc} = \dfrac{2ac}{abc}$

$\\ \implies \sf \dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$

$\\ \implies \sf Hence \: \: proved$

$\\ \rule{220}{2}$