If x²+bc+c=0 and x²+ds+b=0 (b≠c) have a common root, then
1) b+c+1=0
2)b-c+1=0
3)a+c+1=0
4)b+c-1=0
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Answers
Answer:
b + c + 1 = 0
Step-by-step explanation:
Given equation are
x² + bx + c = 0
and
x² + cx + b = 0
Since both quadratic equation have one common root, let that common root be α.
Thus α will satisfy both equations,
α² + bα + c = 0 _______ I)
and
α² + cα + b = 0 _______ II)
Subtracting equation II) from I),
(α² + bα + c) - (α² + cα + b) = 0
α(b - c) + (c - b) = 0
α(b - c) = b - c
α = 1 (∵ b ≠ c)
Putting α = 1 in any of the given quadratic equation,
1² + b·1 + c = 0
1 + b + c = 0
b + c + 1 = 0
Thus option 1) is correct.
Note:- Equation I) and II) are obtained by putting x = α in both quadratic equations respectively.
Answer:
Correct Question :-
- If x² + bx + c = 0 and x² + cx + b = 0, (b ≠ 0) have a common roots, then,
1) b + c + 1 = 0
2) b - c + 1 = 0
3) a + c + 1 = 0
4) b + c - 1 = 0
Given :
- If x² + bx + c = 0 and x² + cx + b = 0, (b ≠ 0) have a common roots.
To Find :-
- What is the equation.
Solution :-
Given equation :
➭ x² + bx + c = 0
➭ x² + cx + b = 0
Now, let α be the common roots.
Then,
➲ α² + bα + c = 0 ---------- (Equation no 1)
➲ α² + cα + b = 0 ---------- (Equation no 2)
Now, we have to subtract equation no (1) from the equation no (2) we get,
↦ α² + bα + c - (α² + cα + b) = 0
↦ α² + bα + c - α² - cα - b = 0
- By cancelling α² from - α² we get,
↦ bα + c - cα - b = 0
↦ bα - cα - b + c = 0
↦ α(b - c) - 1(b - c) = 0
↦ a - 1 = 0
➠ a = 0 since, b ≠ c
Now, by putting α = 1 in the equation no 2 we get,
⇒ α² + cα + b = 0
⇒ (1)² + c.1 + b = 0
⇒ 1 + c + b = 0
- Now, by arranging we get,
➦ b + c + 1 = 0
∴ The equation is b + c + 1 = 0
Hence, the correct options is option no 1) b + c + 1 = 0.