Theory of EquationsmediumFree

Theory of Equations: Root Equation 3ax Value Can Equal

JEE Maths question with a full step-by-step solution.

Question

\tan^2\!\dfrac{3\pi}{8}

is a root of the equation

2x^2 - 3ax + 4b = 0

, where

a, b \in \mathbb{Q}

, then the value of

a + 4b

can be equal to

5

8

6

correct

4

Solution

Step 1:

\tan^2\!\dfrac{3\pi}{8} = \tan^2\!67.5° = (\sqrt{2}+1)^2 = 3 + 2\sqrt{2}

. Step 2: Substitute

x = 3 + 2\sqrt{2}

into

2x^2 - 3ax + 4b = 0

and use

2x^2 = 2(17+12\sqrt{2}) = 34+24\sqrt{2}

(34+24\sqrt{2}) - 3a(3+2\sqrt{2}) + 4b = 0

Step 3: Separating rational and irrational parts (since

a, b \in \mathbb{Q}

): Irrational:

24 - 6a = 0 \implies a = 4

. Rational:

34 - 9a + 4b = 0 \implies 34 - 36 + 4b = 0 \implies b = \dfrac{1}{2}

. Step 4:

a + 4b = 4 + 2 = 6

. Correct answer: (3)

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