Question

if the roots of the equation (a-b)x^2+(b-c)x+(c-a)=0 are equal prove that b+c= 2a​

Answer 1

Answer:

Step-by-step explanation:

Discriminant = (b-c)² - 4(a-b)(c-a)

Since roots are equal discriminant = 0

⇒ b²+c²- 2bc -4(ac-a²-bc+ab) = 0

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

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

⇒ 4a² + b² + c² + 2bc - 4ac - 4ab = 0

⇒ (2a - b - c)² = 0

⇒ 2a - b -c = 0

⇒ 2a = b+c

Answer 2

Given :-

• If the roots of the equations (a - b)x² + (b - c)x + (c - a) = 0 are equal.

To prove :-

• b + c = 2a

Solution :-

D = (b - c)² - 4(a - b) (c - a)

We know that,for real and equal roots we must have D = 0

Now,

$\rm\longrightarrow\:{(b-c)}^{2}-4(a-b)(c-a)=0$

$\rm\longrightarrow\:{4a}^{2}+{b}^{2}+{c}^{2}-4ab+2bc-4ac=0$

$\rm\longrightarrow\:{(-2a)}^{2}+{b}^{2}+{c}^{2}+2(-2a)b+2bc+2c(-2a)=0$

$\rm\longrightarrow\:{(-2a+b+c)}^{2}=0$

$\rm\longrightarrow\:-2a+b+c=0$

$\rm\longrightarrow\:b+c=0+2a$

$\rm\longrightarrow\:b+c=2a$

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Nature of the roots of a quadratic equations :-

Let the given equation be ax² + bx + c = 0 where a ≠ 0 then, the discrimination (D) will be (b² - 4ac)

and roots of the given equation are :-

α = -b + √D /2a and β = -b - √D/2a