The harmonic mean of the roots of the equation (5+√ 2) x2 - (4+√ 5) x + 8 + 2√5 = 0 is

Question

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

Tardigrade Admin · Accepted Answer

Let α, β be the roots of given quadratic equation. then, α+β =(4 +√ 5/5 + √ 2) and αβ =(8 + 2√ 5/5 + √ 2) Let H be the harmonic mean between α and β , then H=(2 αβ/α+β) =(16 + 4√ 5/4 + √ 5)=4

Q. The harmonic mean of the roots of the equation $(5 + 2) x^{2} - (4 + 5) x + 8 + 25 = 0$ is

2

4

6

8

Let $α, β$ be the roots of given quadratic equation. then,
$α + β = \frac{4 + 5}{5 + 2}$ and $α β = \frac{8 + 2 5}{5 + 2}$
Let H be the harmonic mean between $α$ and $β$ , then
$H = \frac{2 α β}{α + β} = \frac{16 + 4 5}{4 + 5} = 4$

Q. The harmonic mean of the roots of the equation (5+2​)x2−(4+5​)x+8+25​=0 is

2

4

6

8

Let α,β be the roots of given quadratic equation. then, α+β=5+2​4+5​​ and αβ=5+2​8+25​​ Let H be the harmonic mean between α and β, then H=α+β2αβ​=4+5​16+45​​=4

Q. The harmonic mean of the roots of the equation $(5 + 2) x^{2} - (4 + 5) x + 8 + 25 = 0$ is

Let $α, β$ be the roots of given quadratic equation. then,
$α + β = \frac{4 + 5}{5 + 2}$ and $α β = \frac{8 + 2 5}{5 + 2}$
Let H be the harmonic mean between $α$ and $β$ , then
$H = \frac{2 α β}{α + β} = \frac{16 + 4 5}{4 + 5} = 4$