Question

aryan0123 please answer this question I will Mark you brainlist​

Answer 1

Answer:

(i) k = 4

(ii) k = 4

(iii) k = 35

Step-by-step explanation:

Concept used:

If one zero is given, then substitute the value of one given zero in the given quadratic equation and then find the value of k.

First part:

f(x) = x² + kx - 21 = 0

Substitute x as 3

f(3) = 3² + 3k - 21 = 0

⇒ f(3) = 9 + 3k - 21 = 0

⇒ 3k - 12 = 0

⇒ 3k = 12

⇒ k = 12 ÷ 3

⇒ k = 4

Second part:

f(x) = x² + kx - 5 = 0

Substitute the value of x as 1

f(1) = 1 + k - 5 = 0

→ k - 4 = 0

→ k = 4

Third part:

f(x) = x² + 12x + k = 0

Substitute the value of x as -5

f(-5) = (-5)² + 12(-5) + k = 0

⇒ 25 - 60 + k = 0

⇒ k - 35 = 0

⇒ k = 35

Answer 2

Find the value of k, such that the number given against each equation is one of it's roots :

$\sf\:(i)\:x^{2}+kx-21=0$ ; $\tt\:3$

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If root of the equation x² + kx - 21 = 0 is 3 then ,

⠀⠀⠀⌬ $\sf\:x~=~3$

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Plugging the value of x as 3

$\sf\implies\:(3)^{2}+k(3)-21=0$

$\sf\implies\:9+3k-21=0$

$\sf\implies\:9+3k=21$

$\sf\implies\:3k=21-9$

$\sf\implies\:3k=12$

$\sf\implies\:k=\frac{12}{3}$

$\sf\implies\:k=\cancel{\frac{12}{3}}$

$\small{\underline{\boxed{\mathrm{\implies\:k=4}}}}$ $\bf\color{red}{⋆}$

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$\sf\:(ii)\:x^{2}+kx-5=0$ ; $\tt\:1$

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If root of the equation x² + kx - 5 = 0 is 1 then ,

⠀⠀⠀⌬ $\sf\:x~=~1$

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Plugging the value of x as 1

$\sf\implies\:(1)^{2}+k(1)-5=0$

$\sf\implies\:1+k-5=0$

$\sf\implies\:1+k=5$

$\sf\implies\:k=5-1$

$\small{\underline{\boxed{\mathrm{\implies\:k=4}}}}$ $\bf\color{red}{⋆}$

════════════════════════

$\sf\:(iii)\:x^{2}+12x+k=0$ ; $\tt\:(-5)$

‎

If root of the equation x² + 12x + k = 0 is (-5) then ,

⠀⠀⠀⌬ $\sf\:x~=~(-5)$

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Plugging the value of x as (-5)

$\sf\implies\:(-5)^{2}+12(-5)+k=0$

$\sf\implies\:25-60+k=0$

$\sf\implies\:25+k=60$

$\sf\implies\:k=60-25$

$\small{\underline{\boxed{\mathrm{\implies\:k=35}}}}$ $\bf\color{red}{⋆}$

════════════════════════

Final AnSwers :

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀(i) k = 4

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀(ii) k = 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀ (iii) k = 35

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