[tex]\huge\sf{\underline{\underline{\red{Q}\green{u}\purple{e}\pink{s}\blue{t}\orange{i}\purple{o}\green{n}}}}[/tex]
If k > 0 and the product of the roots of the equation [tex]\sf{x² - 3kx +{2e}^{2logk} -1=0}[/tex]is 7 then the sum of the roots is
(a) 2
(b) 4
(c) 6
(d) 8
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Since this question was asked by me once, I can readily give this answer.... (θ‿θ)
So,
Given the equation :
[tex]\sf{x² - 3kx +{2e}^{2logk} -1=0}[/tex]
[tex]\tt{a \: = \: 1 \: \: b \: = \: 3k \: \: c = {2e}^{2logk} \: -\: 1}[/tex]
Now,
Given that the product of the equation is 7,
A/Q,
[tex]\sf{{2e}^{2logk}-1 \: = \: 7}[/tex]
[tex]\sf{{2e}^{2logk} \: = \: 8}[/tex]
[tex]\sf{{e}^{2logk}\: = \ 4}[/tex]
[tex]\sf{{e}^{log{k}^{2}} \: = \: 4}[/tex]
[tex]\sf{{k}^{2}\: = \: 4}[/tex]
[tex]\sf{k \: = \: ±2}[/tex]
Since, k < 0,
[tex]\leadsto{ k \: = \: -2}[/tex]
Now, sum of roots = -3k
[tex]\dashrightarrow{-3(-2)}[/tex]
[tex]\longrightarrow{\pink{6}}[/tex]
Have a furr-tastic day ahead
Verified answer
[tex]\huge{\bf{{\underline{\colorbox{black} {\color{white}{➳Aɳʂɯҽɾ:-}}}}}}[/tex]
To find the sum of the roots of the quadratic equation, we can use Vieta's formulas. Let's start by expressing the quadratic equation in the standard form:
x² - 3kx + 2e²log(k) - 1 = 0
The general form of a quadratic equation is ax² + bx + c = 0, so we can compare the coefficients of the given equation to identify the values of a, b, and c.
a = 1
b = -3k
c = 2e²log(k) - 1
α + β = -b/a
α + β = -(-3k) / 1
α + β = 3k
Given that the product of the roots is 7, we can use another Vieta's formula, which states that the product of the roots is equal to the constant term divided by the leading coefficient:
αβ = c/a
αβ = (2e²log(k) - 1) / 1
αβ = 2e²log(k) - 1
2e²log(k) - 1 = 7
2e²log(k) = 8
log(k) = 4 / (2e²)
k = e^(4/(2e²))
α + β = 3(e^(4/(2e²)))
To determine the exact sum, we could simplify further if needed, but based on the answer choices provided, it appears that the exact value is not required.
Therefore, the sum of the roots is given by option (d) 8.
[tex]\huge\rm\pmb{Hi...}[/tex]