The harmonic mean of the roots of the equatio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the quadratic equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$.Let the roots be $x_1$ and $x_2$.From the properties of qua

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the quadratic equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$.Let the roots be $x_1$ and $x_2$.From the properties of quadratic equations, the sum of roots is $x_1 + x_2 = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$.The product of roots is $x_1 x_2 = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}} = \frac{2(4 + \sqrt{5})}{5 + \sqrt{2}} = 2(x_1 + x_2)$.The harmonic mean $(HM)$ of two numbers $x_1$ and $x_2$ is given by $HM = \frac{2x_1 x_2}{x_1 + x_2}$.Substituting $x_1 x_2 = 2(x_1 + x_2)$ into the formula, we get $HM = \frac{2 \cdot 2(x_1 + x_2)}{x_1 + x_2} = 4$.

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

Similar Questions

Solve the given two equations and select the correct option.
$I.$ $77 x^{2}+58 x+8=0$
$II.$ $42 y^{2}+59 y+20=0$

If $\alpha, \beta$ and $\gamma$ are the roots of $x^3 + px + q = 0$,then the value of $\alpha^3 + \beta^3 + \gamma^3$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

Let $p(x)$ be a quadratic polynomial such that $p(0) = 1$. If $p(x)$ leaves a remainder of $4$ when divided by $x - 1$ and a remainder of $6$ when divided by $x + 1$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-11x+24=0$
$II.$ $2y^{2}-9y+9=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module