[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
[tex]\huge\tt\color{beige}{❀⊱┄┄┄┄⊰❀}[/tex]
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[tex]\sf\green{No \: ChatGPT \: or \: Bard \: Copied \: Answers}[/tex]
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Thanks A Lot
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
[tex]\huge\tt\color{beige}{❀⊱┄┄┄┄⊰❀}[/tex]
Only Genius, Moderators and Brainly Stars will Answer
No one Else
It's a Request
Dont SPAM
[tex]\sf\green{No \: ChatGPT \: or \: Bard \: Copied \: Answers}[/tex]
[tex]\huge\tt\color{beige}{❀⊱┄┄┄┄⊰❀}[/tex]
Thanks A Lot
Answers & Comments
Verified answer
Answer:
[tex]\sf{x² - 3kx +{2e}^{2logk} -1=0}[/tex]
[tex]a = 1, b = -3k, c = 2 {e}^{2log \: k} - 1[/tex]
product of roots = 7
[tex] \frac{c}{a} = 7[/tex]
[tex] \frac{2 {e}^{2log \: k} - 1}{1} = 7[/tex]
[tex] 2 {e}^{2log \: k} - 1 = 7[/tex]
[tex] 2 {e}^{2log \: k} = 7 + 1[/tex]
[tex] 2 {e}^{2log \: k} = 8[/tex]
divide by 2 on both sides,
[tex] {e}^{2log \: k} = 4[/tex]
[tex]alog \: b = log {b}^{a} [/tex]
[tex] {e}^{log \: {k}^{2} } = 4[/tex]
k² = 4
k = ±2
given k > 0, k = 2
[tex]sum \: of \: roots = \frac{ - b}{a} [/tex]
[tex]sum \: of \: roots = \frac{ - ( - 3k)}{1} [/tex]
sum of roots = 3k = 3(2) = 6
option C)