Question

if one root of x^2+kx+12=0 may be triple of the other , then k is ​

Answer 1

Answer:

${x}^{2} + kx + 12 = 0$

Let α, β be the roots

∴β = 3α → (condition given)

a = 1 , b = k , c = 12

sum of the roots

$α + β = \frac{ - b}{a}$

$α + 3α = \frac{ - k}{1}$

$4α = - k$

$α = \frac{ - k}{4}$

product of the roots

$α \times β \: = \frac{c}{a}$

$α(3α) = \frac{12}{1}$

$3α^{2} = 12$

$α^{2} = \frac{12}{3} = 4$

$α = 2$

comparing sum and product α, β

$\implies\frac{ - k}{4} = 2$

$\implies - k = 4 \times 2$

$\boxed {\implies \: \: k = - 8}$

Answer 2

Answer:

Answer is k = 8 , -8 .

Step-by-step explanation:

Given :- $x^2+kx+12 = 0$

To find :- value of k .

Solution :-

Step 1) Let , α and β be the root of equation .

Step 2) product of roots , α*β = c/a --- (2) , α + β = -b/a ---- (3) .

Step 3) Given condition is that , the α = 3*β -- (4) .

therefore ,

α + β = -k/1

3β + β = -k

4β = -k --- (5)

Step 4)

Product of roots , α*β = 12/1

3β*β=12 /1

$3\beta ^2 = 12 \\\beta ^2 = 4 \\\beta = +_- 2$

Step 5)

• If β = 2 , k = -8
• If β = -2 , k = 8 .

Therefore , answer is k = 8 , -8 .

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