Question

if tbe roots of the quadratic equation (a-b)×2+(b-c )x +(c-a)=0are equal then realtion between a,b and c​

Answer 1

$\huge{\underbrace{ \mathfrak {\blue{Answer}}}}$

The question must be, if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.

Using Discriminant,

D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0

so, A = a-b

B = b-c

C = c-a

For roots to be equal, D=0

(b-c)2 - 4(a-b)(c-a) =0

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

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

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

(2a-b-c)2=0

i.e. 2a-b-c =0

2a= b+c

Answer 2

Given :-

• If tbe roots of the quadratic equation (a-b)×2+(b-c )x +(c-a)=0 are equal

To find :-

• Relation between a,b and c

Solution :-

Now, as we know that

• b² - 4ac > 0 → two distinct real roots
• b² - 4ac = 0 → two equal real roots
• b² - 4ac < 0 → no real roots

✶ In above question it is given that quadratic equation has equal roots.

According to quadratic formula

→ b² - 4ac = 0 (equal roots)

• (a - b)x² + (b - c)x + (c - a) [Given equation]
• a = (a - b)
• b = (b - c)
• c = (c - a)

Now, substitute the value of a,b and c

→ (b - c)² - 4 × (a - b) × (c - a) = 0

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

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

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

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

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

→ 4a² + b² + c² - 4ab + 2bc - 4ac = 0

Apply identity : (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

→ (-2a)² + (b)² + (c)² + 2 × (-2a) × b + 2 × b × c + 2 × (-2a) × c = 0

→ (-2a + b + c)² = 0

→ - 2a + b + c = 0

→ 2a = b + c

Hence,

• Relation between a,b and c i.e 2a = b + c

Note :

A quadratic equation is in the form ax² + bc + c = 0, where a,b,c are real number and a ≠ 0