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Given:
- x² + kx - 21 = 0; its root = 3
- x² + kx - 5 = 0; its root = 1
- x² + 12x + k = 0; its root = -5
To find: The value of k.
Answer:
We're given both the equations and their roots. To find the value of k, substitute the value of x with the respective root.
1. x² + kx - 21 = 0
- x = 3
3² + (k × 3) - 21 = 0
9 + 3k - 21 = 0
-12 + 3k = 0
3k = 12
k = 12 ÷ 4
k = 3
2. x² + kx - 5 = 0
- x = 1
1² + (k × 1) - 5 = 0
1 + k - 5 = 0
-4 + k = 0
k = 4
3. x² + 12x + k = 0
- x = -5
-5² + 12 × (-5) + k = 0
25 - 60 + k = 0
-35 + k = 0
k = 35
Therefore, the values of k for each of the equations are 3, 4 and 35.
Given :
(I) x² + kx - 21 = 0
(II) x² + kx - 5 = 0
(III) x² + 12x + k = 0
To find :
The value of k, such that the number given against each equation is one of its roots.
(I) 3 (II) 1 (III) -5
Solution :
(I) f(x) = x² + kx - 21 = 0
We have, x = 3
⇒ 3² + 3k - 21 = 0
⇒ 9 + 3k - 21 = 0
⇒ 3k - 12 = 0
⇒ 3k = 12
⇒ k = 12/3
⇒ k = 4
(II) f(x) = x² + kx - 5 = 0
We have, x = 1
⇒ 1 + k - 5 = 0
⇒ k - 4 = 0
⇒ k = 4
(III) f(x) = x² + 12x + k = 0
We have, x = -5
⇒ (-5)² + 12(-5) + k = 0
⇒ 25 + (-60) + k = 0
⇒ 25 - 60 + k = 0
⇒ k - 35 = 0
⇒ k = 35
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