Multiple choice

If the product of the roots of the equation $x^2-3kx+2e^{2:log:k}-1=0$ is $7$, then the roots of the equation are real if $k$ equals-

  1. $1$
  2. $3$
  3. $-4$
  4. $\pm2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Product of roots = c/a = 2e^(2 log k) - 1 = 7. Since e^(log k^2) = k^2, we have 2k^2 - 1 = 7, so 2k^2 = 8, k^2 = 4, k = +/- 2. For real roots, discriminant D = (-3k)^2 - 4(1)(7) >= 0, so 9k^2 - 28 >= 0. With k^2 = 4, 9(4) - 28 = 36 - 28 = 8 > 0, so roots are real.

AI explanation

Using the property of the product of roots for x squared minus 3kx plus 2e raised to the power of 2 log k minus 1 equals 0, we set 2e raised to the power of 2 log k minus 1 equal to 7. This simplifies to e raised to the power of 2 log k equals 4, so 2 log k equals log 4, meaning k squared equals 4 and k equals plus or minus 2. Checking the discriminant D equals 9k squared minus 28, we get D equals 36 minus 28, which is 8 (a positive number), confirming the roots are real. Therefore, k equals plus or minus 2.

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