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