If k > 0 and the Product of the Roots of the Equation x² − 3kx + 2e^(2 log k) − 1 = 0 Is 7, Then the Sum of the Roots Is

A standard JEE-style quadratic equations problem.

Problem

If $k>0$ and the product of the roots of the equation $x^2-3kx+2e^{2\log k}-1=0$ is $7$, then the sum of the roots is:

Note: the working in the solution embedded below reads the constant term as $2e^{\log k}-1$, which gives a sum of $12$, not one of the options. The standard form of this question has $2e^{2\log k}-1$, which gives option (D), and that form is used here.

Key insight. The constant term $2e^{2\log k}-1$ looks intimidating, but $e^{2\log k}=e^{\log k^2}=k^2$ — the exponential and logarithm simply undo each other once the exponent $2$ is absorbed inside the logarithm. Once that’s simplified, the whole problem is just Vieta’s formulas applied to an ordinary quadratic in $x$, with $k$ as a parameter to solve for first.

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Approach

Simplify the constant term $2e^{2\log k}-1$ into a plain expression in $k$. Use Vieta’s formula for the product of roots to form an equation in $k$ and solve for it (keeping only the value consistent with $k>0$). Then use Vieta’s formula for the sum of roots, now that $k$ is known.

Solution

Step 1 — Simplify the constant term

$$e^{2\log k} = e^{\log k^2} = k^2$$

so the equation is:

$$x^2-3kx+(2k^2-1)=0$$

Step 2 — Use the product-of-roots formula

For $x^2+bx+c=0$, the product of the roots is $c$ (here $a=1$). Given the product is $7$:

$$2k^2-1=7$$

Step 3 — Solve for k

$$2k^2=8 \implies k^2=4 \implies k=\pm2$$

Since $k>0$:

$$k=2$$

Step 4 — Use the sum-of-roots formula

For $x^2+bx+c=0$, the sum of the roots is $-b$ (here $a=1$). The coefficient of $x$ is $-3k$, so the sum of the roots is $3k$:

$$\text{Sum} = 3k = 3(2) = 6$$

Answer

Common mistakes

• Simplifying $e^{2\log k}$ as $2k$ or $e^{\log k}\cdot2$ instead of $k^2$. The exponent $2$ multiplies the logarithm, so it has to be absorbed as a power before undoing the log: $e^{2\log k}=e^{\log(k^2)}=k^2$, not $2\cdot e^{\log k}=2k$.
• Forgetting the condition $k>0$ and keeping both roots of $k^2=4$. $k=-2$ would also satisfy $2k^2-1=7$ algebraically, but $\log k$ is undefined for negative $k$, so only $k=2$ is valid here.
• Reading off the sum of roots as $-3k$ instead of $3k$. For $x^2+bx+c=0$ the sum of roots is $-b$; here $b=-3k$, so the sum is $-(-3k)=3k$ — dropping the double negative gives the wrong sign.

Practise next

• If $m>0$ and the product of the roots of the equation $x^2-4mx+3e^{2\log m}-2=0$ is $10$, find the sum of the roots.