The harmonic mean of the roots of the equation (2+√3)x2-(3+√5)x+(6+2√5)=0 is

Question

The harmonic mean of the roots of the equation (2+√3)x2-(3+√5)x+(6+2√5)=0 is (A) 2 (B) 7 (C) 8 (D) 4

Tardigrade Admin · Accepted Answer

Given equation is (2+√3)x2-(3+√5)x+(6+2√5)=0 let α, β be its roots ∴ α+β=(3+√5/2+√3) and αβ=(6+2√5/2+√3) ∴ H.M. of α and β is given by H=(2αβ/α+β)=(2(6+2√5)/2+√3)×(2+√3/3+√5)=4(1)=4

Q. The harmonic mean of the roots of the equation $(2 + 3) x^{2} - (3 + 5) x + (6 + 25) = 0$ is

$2$

$7$

$8$

$4$

Given equation is
$(2 + 3) x^{2} - (3 + 5) x + (6 + 25) = 0$
let $α, β$ be its roots
$∴ α + β = \frac{3 + 5}{2 + 3}$ and $α β = \frac{6 + 2 5}{2 + 3}$
$∴$ H.M. of $α$ and $β$ is given by
$H = \frac{2 α β}{α + β} = \frac{2 ( 6 + 2 5 )}{2 + 3} \times \frac{2 + 3}{3 + 5} = 4 (1) = 4$

Q. The harmonic mean of the roots of the equation (2+3​)x2−(3+5​)x+(6+25​)=0 is

2

7

8

4

Given equation is (2+3​)x2−(3+5​)x+(6+25​)=0 let α,β be its roots ∴α+β=2+3​3+5​​ and αβ=2+3​6+25​​ ∴ H.M. of α and β is given by H=α+β2αβ​=2+3​2(6+25​)​×3+5​2+3​​=4(1)=4

Q. The harmonic mean of the roots of the equation $(2 + 3) x^{2} - (3 + 5) x + (6 + 25) = 0$ is

$2$

$7$

$8$

$4$

Given equation is
$(2 + 3) x^{2} - (3 + 5) x + (6 + 25) = 0$
let $α, β$ be its roots
$∴ α + β = \frac{3 + 5}{2 + 3}$ and $α β = \frac{6 + 2 5}{2 + 3}$
$∴$ H.M. of $α$ and $β$ is given by
$H = \frac{2 α β}{α + β} = \frac{2 ( 6 + 2 5 )}{2 + 3} \times \frac{2 + 3}{3 + 5} = 4 (1) = 4$