If the harmonic mean of the roots of the equa | vedclass

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

Question

(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. From the properties

Vedclass · Accepted Answer

$4 - \sqrt{5}$

Vedclass · Answer

(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = \frac{-(-b)}{\sqrt{2}} = \frac{b}{\sqrt{2}}$. The product of the roots is $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}}$. The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$. Given $HM = 4$,we have: $\frac{2\alpha\beta}{\alpha + \beta} = 4$ Substituting the values of $\alpha + \beta$ and $\alpha\beta$: $\frac{2 \left( \frac{8 - 2\sqrt{5}}{\sqrt{2}} \right)}{\frac{b}{\sqrt{2}}} = 4$ $\frac{2(8 - 2\sqrt{5})}{b} = 4$ $2(8 - 2\sqrt{5}) = 4b$ $8 - 2\sqrt{5} = 2b$ $b = 4 - \sqrt{5}$.

If the harmonic mean of the roots of the equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is:

Similar Questions

If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2 > AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

Bhutia is a breed of:

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$,then $k$ is equal to

If $y$ is a function of $x$ and $\log(x+y)=2xy$,then the value of $y'(0)$ is

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module