If tan 2 (3 π/8) is a root of the equation 2 x 2-3 ax +4 b =0, where a, b ∈ Q, then the value of a+4 b can be equal to:

Question

If tan 2 (3 π/8) is a root of the equation 2 x 2-3 ax +4 b =0, where a, b ∈ Q, then the value of a+4 b can be equal to: (A) 5 (B) 8 (C) 6 (D) 4

Tardigrade Admin · Accepted Answer

tan 2 (3 π/8)=(√2+1)2=3+2 √2 ∴ x=3+2 √2 satisfy the equation 2 x2-3 a x+4 b=0... (1) (x-3)2=8 ⇒ x2-6 x+1=0...(2) comparing (1) and (2), we get ∴ a =4 and b =1 / 2 ⇒ a +4 b =6

Q. If $tan^{2} \frac{3 π}{8}$ is a root of the equation $2 x^{2} - 3 a x + 4 b = 0$ , where $a, b \in Q$ , then the value of $a + 4 b$ can be equal to:

5

8

6

4

$tan^{2} \frac{3 π}{8} = (2 + 1)^{2} = 3 + 22$
$∴ x = 3 + 22$ satisfy the equation
$2 x^{2} - 3 a x + 4 b = 0...$ (1)
$(x - 3)^{2} = 8 \Rightarrow x^{2} - 6 x + 1 = 0...$ (2)
comparing (1) and (2), we get
$∴ a = 4$ and $b = 1/2$
$\Rightarrow a + 4 b = 6$

Q. If tan283π​ is a root of the equation 2x2−3ax+4b=0, where a,b∈Q, then the value of a+4b can be equal to:

5

8

6

4

tan283π​=(2​+1)2=3+22​ ∴x=3+22​ satisfy the equation 2x2−3ax+4b=0...(1) (x−3)2=8⇒x2−6x+1=0...(2) comparing (1) and (2), we get ∴a=4 and b=1/2 ⇒a+4b=6

Q. If $tan^{2} \frac{3 π}{8}$ is a root of the equation $2 x^{2} - 3 a x + 4 b = 0$ , where $a, b \in Q$ , then the value of $a + 4 b$ can be equal to:

$tan^{2} \frac{3 π}{8} = (2 + 1)^{2} = 3 + 22$
$∴ x = 3 + 22$ satisfy the equation
$2 x^{2} - 3 a x + 4 b = 0...$ (1)
$(x - 3)^{2} = 8 \Rightarrow x^{2} - 6 x + 1 = 0...$ (2)
comparing (1) and (2), we get
$∴ a = 4$ and $b = 1/2$
$\Rightarrow a + 4 b = 6$