JEE MainMathematics

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Q: Why do we divide by t^2 and use u = t - 1/t instead of u = t + 1/t?

The coefficients of the cubic and linear terms are +8 and -8. Factoring out 8 gives 8(t - 1/t), which dictates using u = t - 1/t.

Q: Why does 0 < t < 1 mean x is negative?

Here, . The natural logarithm function is strictly increasing, and . Therefore, is strictly negative when .

The equation

\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}

has :

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Afour solutions two of which are negativeBtwo solutions and only one of them is negativeCtwo solutions and both are negativeDno solution

Step-by-step solutionView

Correct answer

Substituting $t = \mathrm{e}^x$ and dividing by $t^2$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $t$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Option analysis

Why each option works or fails

A · four solutions two of which are negative

The student treats all four roots of the quartic polynomial in $t = \mathrm{e}^x$ as valid real values for $x$ , forgetting that $t$ must be strictly positive. Check the condition $t > 0$ for $t = \mathrm{e}^x$ ; negative values of $t$ cannot produce real values of $x$ .

B · two solutions and only one of them is negative

The student identifies two positive roots for $t = \mathrm{e}^x$ but assumes that one root is greater than $1$ (yielding a positive $x$ ) and one is less than $1$ (yielding a negative $x$ ). Compare each positive root $t$ to $1$ : because both satisfy $0 < t < 1$ , both corresponding solutions $x = \ln(t)$ are strictly negative.

C · two solutions and both are negative

None. The student correctly transforms the equation, enforces $t = \mathrm{e}^x > 0$ , and checks whether $t > 1$ or $0 < t < 1$ . Correctly observed that the substitution gives two valid roots for $t$ , both lying in the interval $(0, 1)$ , giving two negative values for $x$ .

D · no solution

The student makes an algebraic sign error when substituting $u = t - 1/t$ or $u = t + 1/t$ , concluding that the quadratic has no real roots. Carefully square $t - 1/t = u$ to get $t^2 + 1/t^2 = u^2 + 2$ , and solve the resulting quadratic $u^2 + 8u + 15 = 0$ .

Reviewed route

Solution

StepWorking

01given

The equation is $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$ for $x \in \mathbb{R}$ .

02goal

Determine the number of real solutions $x$ and their signs.

03approach

Substitute $t = e^x > 0$ . Since the polynomial $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ has anti-symmetric inner coefficients ( $+8$ and $-8$ ), divide by $t^2$ and use the auxiliary substitution $u = t - \frac{1}{t}$ . Solve the resulting quadratic in $u$ , find the corresponding positive roots $t$ , and deduce the signs of $x = \ln t$ .

04execute

Substitute $t = e^x$ with $t > 0$ : $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ Dividing by $t^2 \neq 0$ : $\left(t^2 + \frac{1}{t^2}\right) + 8\left(t - \frac{1}{t}\right) + 13 = 0$ Let $u = t - \frac{1}{t}$ . Then $u^2 = t^2 - 2 + \frac{1}{t^2} \implies t^2 + \frac{1}{t^2} = u^2 + 2$ . The equation becomes: $(u^2 + 2) + 8u + 13 = 0 \implies u^2 + 8u + 15 = 0$ Factoring gives: $(u + 3)(u + 5) = 0 \implies u = -3 \text{ or } u = -5$

05execute

Now solve for $t > 0$ from each value of $u$ : Case 1: $t - \frac{1}{t} = -3 \implies t^2 + 3t - 1 = 0$ The roots are $t = \frac{-3 \pm \sqrt{9 - 4(1)(-1)}}{2} = \frac{-3 \pm \sqrt{13}}{2}$ . Since $t > 0$ , we have one valid root: $t_1 = \frac{\sqrt{13} - 3}{2}$ . Since $3 = \sqrt{9} < \sqrt{13} < \sqrt{25} = 5$ , we have $0 < \sqrt{13} - 3 < 2$ , so $0 < t_1 < 1$ . Case 2: $t - \frac{1}{t} = -5 \implies t^2 + 5t - 1 = 0$ The roots are $t = \frac{-5 \pm \sqrt{25 - 4(1)(-1)}}{2} = \frac{-5 \pm \sqrt{29}}{2}$ . Since $t > 0$ , we have one valid root: $t_2 = \frac{\sqrt{29} - 5}{2}$ . Since $5 = \sqrt{25} < \sqrt{29} < \sqrt{49} = 7$ , we have $0 < \sqrt{29} - 5 < 2$ , so $0 < t_2 < 1$ .

06execute

Since $x = \ln t$ , for both roots we have $0 < t < 1$ , which gives $\ln t < 0$ . Thus, $x_1 = \ln t_1 < 0$ and $x_2 = \ln t_2 < 0$ . Therefore, there are exactly two real solutions, and both are negative.

✓verify

Each quadratic $t^2 + ku - 1 = 0$ has constant term $-1$ , meaning its two roots have product $-1$ . Hence, exactly one root is positive and one is negative for each equation, yielding precisely two positive values of $t$ . Since $t^2 + kt - 1 = 0$ evaluated at $t=1$ gives $k$ , and for $k \in \{3, 5\}$ we have $1 + k - 1 = k > 0$ while at $t=0$ it is $-1 < 0$ , the single positive root strictly lies in $(0, 1)$ . Thus $\ln t < 0$ for both, confirming both solutions $x$ are negative.

Hints that build this answer step by step

Letting $t = \mathrm{e}^x > 0$ , the equation becomes $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ . What is the most effective algebraic reduction?
Divide by $t^2$ and substitute $u = t - \frac{1}{t}$
Using $u = t - \frac{1}{t}$ , we have $t^2 + \frac{1}{t^2} = u^2 + 2$ . What are the values of $u$ ?
$u = -3$ and $u = -5$
For $u = -3$ and $u = -5$ , how many positive solutions exist for $t$ in $t - \frac{1}{t} = u$ , and where do they lie relative to $1$ ?
Each value of $u$ gives exactly one positive solution $t$ , and both satisfy $0 < t < 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we divide by t^2 and use u = t - 1/t instead of u = t + 1/t?

The coefficients of the cubic and linear terms are +8 and -8. Factoring out 8 gives 8(t - 1/t), which dictates using u = t - 1/t.

Why does 0 < t < 1 mean x is negative?

Here, $x = \ln(t)$ . The natural logarithm function is strictly increasing, and $\ln(1) = 0$ . Therefore, $\ln(t)$ is strictly negative when $0 < t < 1$ .

Answer

Substituting $t = \mathrm{e}^x$ and dividing by $t^2$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Why each option works or fails

A: four solutions two of which are negative - The student treats all four roots of the quartic polynomial in $t = \mathrm{e}^x$ as valid real values for $x$ , forgetting that $t$ must be strictly positive. Check the condition $t > 0$ for $t = \mathrm{e}^x$ ; negative values of $t$ cannot produce real values of $x$ .
B: two solutions and only one of them is negative - The student identifies two positive roots for $t = \mathrm{e}^x$ but assumes that one root is greater than $1$ (yielding a positive $x$ ) and one is less than $1$ (yielding a negative $x$ ). Compare each positive root $t$ to $1$ : because both satisfy $0 < t < 1$ , both corresponding solutions $x = \ln(t)$ are strictly negative.
C · correct: two solutions and both are negative - None. The student correctly transforms the equation, enforces $t = \mathrm{e}^x > 0$ , and checks whether $t > 1$ or $0 < t < 1$ . Correctly observed that the substitution gives two valid roots for $t$ , both lying in the interval $(0, 1)$ , giving two negative values for $x$ .
D: no solution - The student makes an algebraic sign error when substituting $u = t - 1/t$ or $u = t + 1/t$ , concluding that the quadratic has no real roots. Carefully square $t - 1/t = u$ to get $t^2 + 1/t^2 = u^2 + 2$ , and solve the resulting quadratic $u^2 + 8u + 15 = 0$ .

Step-by-step solution

given: The equation is $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$ for $x \in \mathbb{R}$ .
goal: Determine the number of real solutions $x$ and their signs.
approach: Substitute $1$ . Since the polynomial $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ has anti-symmetric inner coefficients ( $+8$ and $-8$ ), divide by $t^2$ and use the auxiliary substitution $u = t - \frac{1}{t}$ . Solve the resulting quadratic in $u$ , find the corresponding positive roots $t$ , and deduce the signs of $x = \ln t$ .
execute: Substitute $t = e^x$ with $t > 0$ : $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ Dividing by $t^2 \neq 0$ : $\left(t^2 + \frac{1}{t^2}\right) + 8\left(t - \frac{1}{t}\right) + 13 = 0$ Let $u = t - \frac{1}{t}$ . Then $u^2 = t^2 - 2 + \frac{1}{t^2} \implies t^2 + \frac{1}{t^2} = u^2 + 2$ . The equation becomes: $(u^2 + 2) + 8u + 13 = 0 \implies u^2 + 8u + 15 = 0$ Factoring gives: $(u + 3)(u + 5) = 0 \implies u = -3 \text{ or } u = -5$
execute: Now solve for $t > 0$ from each value of $u$ : Case 1: $t - \frac{1}{t} = -3 \implies t^2 + 3t - 1 = 0$ The roots are $t = \frac{-3 \pm \sqrt{9 - 4(1)(-1)}}{2} = \frac{-3 \pm \sqrt{13}}{2}$ . Since $t > 0$ , we have one valid root: $t_1 = \frac{\sqrt{13} - 3}{2}$ . Since $3 = \sqrt{9} < \sqrt{13} < \sqrt{25} = 5$ , we have $0 < \sqrt{13} - 3 < 2$ , so $0 < t_1 < 1$ . Case 2: $t - \frac{1}{t} = -5 \implies t^2 + 5t - 1 = 0$ The roots are $t = \frac{-5 \pm \sqrt{25 - 4(1)(-1)}}{2} = \frac{-5 \pm \sqrt{29}}{2}$ . Since $t > 0$ , we have one valid root: $t_2 = \frac{\sqrt{29} - 5}{2}$ . Since $5 = \sqrt{25} < \sqrt{29} < \sqrt{49} = 7$ , we have $0 < \sqrt{29} - 5 < 2$ , so $0 < t_2 < 1$ .
execute: Since $x = \ln t$ , for both roots we have $0 < t < 1$ , which gives $\ln t < 0$ . Thus, $x_1 = \ln t_1 < 0$ and $x_2 = \ln t_2 < 0$ . Therefore, there are exactly two real solutions, and both are negative.
verify: Each quadratic $t^2 + ku - 1 = 0$ has constant term $-1$ , meaning its two roots have product $-1$ . Hence, exactly one root is positive and one is negative for each equation, yielding precisely two positive values of $t$ . Since $t^2 + kt - 1 = 0$ evaluated at $t=1$ gives $k$ , and for $k \in \{3, 5\}$ we have $1 + k - 1 = k > 0$ while at $t=0$ it is $-1 < 0$ , the single positive root strictly lies in $(0, 1)$ . Thus $\ln t < 0$ for both, confirming both solutions $x$ are negative.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

Substituting $t = \mathrm{e}^x$ and dividing by $t^2$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $t$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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