Theory of EquationsmediumFree

Theory of Equations: Root Equation 3ax Value Can Equal

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

Question
If tan⁡2 ⁣3π8\tan^2\!\dfrac{3\pi}{8} is a root of the equation 2x2−3ax+4b=02x^2 - 3ax + 4b = 0, where a,b∈Qa, b \in \mathbb{Q}, then the value of a+4ba + 4b can be equal to
A55
B88
C66correct
D44
Solution
Step 1: tan⁡2 ⁣3π8=tan⁡2 ⁣67.5°=(2+1)2=3+22\tan^2\!\dfrac{3\pi}{8} = \tan^2\!67.5° = (\sqrt{2}+1)^2 = 3 + 2\sqrt{2}. Step 2: Substitute x=3+22x = 3 + 2\sqrt{2} into 2x2−3ax+4b=02x^2 - 3ax + 4b = 0 and use 2x2=2(17+122)=34+2422x^2 = 2(17+12\sqrt{2}) = 34+24\sqrt{2}:
(34+242)−3a(3+22)+4b=0(34+24\sqrt{2}) - 3a(3+2\sqrt{2}) + 4b = 0
 Step 3: Separating rational and irrational parts (since a,b∈Qa, b \in \mathbb{Q}): Irrational: 24−6a=0  ⟹  a=424 - 6a = 0 \implies a = 4. Rational: 34−9a+4b=0  ⟹  34−36+4b=0  ⟹  b=1234 - 9a + 4b = 0 \implies 34 - 36 + 4b = 0 \implies b = \dfrac{1}{2}. Step 4: a+4b=4+2=6a + 4b = 4 + 2 = 6. Correct answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Theory of Equations · easy
The values of α\alpha and β\beta for which the quadratic equation x2+2x+2+e2α−cos⁡β=0x^2 + 2x + 2 + e^{2\alpha} - \cos\beta = 0 has a real solution is
Theory of Equations · easy
If the vertex of the quadratic expression y=2x2−4x+6y = 2x^2 - 4x + 6 is (m,n)(m, n), then the quadratic equation whose vertex is (2m,n2)\left(2m, \dfrac{n}{2}\right) is
Theory of Equations · easy
If α,β\alpha, \beta are the roots of x2−p(x+1)−c=0x^2 - p(x+1) - c = 0, c≠1c \neq 1, then the value of (α+1)(β+1)(\alpha+1)(\beta+1) is equal to
Theory of Equations · easy
Let α\alpha and β\beta be the roots of the equation x2−(p−2)x−(p−1)=0x^2 - (p-2)x - (p-1) = 0, p∈Rp \in \mathbb{R}. If α2+β2\alpha^2 + \beta^2 is least, then pp equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath