Question

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please answer this question please​

Answer 1

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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