Question

if the roots of the equation a minus b into X square + b minus C Into X + c minus A is equal to zero are equal prove that b + C is equal to 2a​

Answer 1

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Answer 2

Question

If the roots of the equation (a - b)x² + (b - c)x + (c - a) = 0 are equal prove that b + c = 2a

To Prove

b + c = 2a

Proof

Given equation is, (a - b)x² + (b - c)x + (c - a) = 0

Also, given that roots of the above equation are equal.

If roots are equal then, D = b² - 4ac = 0

We have, a = (a - b), b = (b - c) and c = (c - a)

Substitute the known values in above formula

→ b² - 4ac = 0

→ (b - c)² - 4(a - b)(c - a) = 0

→ b² + c² - 2bc -4[a(c - a) -b(c - a)] = 0

Used identity: (a - b)² = a² + b² - 2ab

→ b² + c² - 2bc -4(ac - a² - bc + ab) = 0

Solve the brackets

→ b² + c² - 2bc - 4ac + 4a² + 4bc - 4 ab = 0

→ 4a² + b² + c² - 4ab - 2bc + 4bc - 4ac = 0

→ 4a² + b² + c² - 4ab + 2bc - 4ac = 0

Also, we can write 4a² as (2a)², 4ab as 2(2ab) and 4ac as 2(2ac)

→ (2a)² + b² + c² - 2(2ab) + 2bc - 2(2ac) = 0

We know that (-2a + b + c)² = (-2a)² + (b)² + (c)² - 2(-2a)(b) - 2(b)(c) - 2(c)(-2a)

= 4a² + b² + c² + 4ab - 2bc + 4ac

Used identity: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

→ (-2a + b + c)² = 0

→ - 2a + b + c = 0

→ b + c = 2a

Hence, proved.

We can also solve it like,

→ 4a² + b² + c² - 4ab - 2bc + 4bc - 4ac = 0

→ b² + c² - 2bc = - 4a² + 4ab + 4ac

→ (b + c)² = - 4a² + 4ab + 4ac

→ (b + c)² = - 4a² + 4a(b + c)

→ (b + c)² = - 4a² + 2a(b + c) + 2a(b + c)

→ (b + c)² - 2a(b + c) = 2a(-2a + b + c)

Take (b + c) common on LHS

→ (b + c)[(b + c) - 2a] = 2a(-2a + b + c)

→ (b + c)(b + c - 2a) = 2a(-2a + b + c)

(-2a + b + c) cancel throughout

→ b + c = 2a

Hence, proved