The harmonic mean of two numbers is -frac{8}{ | vedclass

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

Question

(B) Let the two numbers be $a$ and $b$.Given,Geometric Mean ($G$.$M$.) $= \sqrt{ab} = 2$,so $ab = 4$ (Eq. $i$).Harmonic Mean ($H$.$M$.) $= \frac{2ab}{

Vedclass · Accepted Answer

$x^2+10x+16=0$

Vedclass · Answer

(B) Let the two numbers be $a$ and $b$.Given,Geometric Mean ($G$.$M$.) $= \sqrt{ab} = 2$,so $ab = 4$ (Eq. $i$).Harmonic Mean ($H$.$M$.) $= \frac{2ab}{a+b} = -\frac{8}{5}$.Substituting $ab = 4$ into the $H$.$M$. formula:$\frac{2(4)}{a+b} = -\frac{8}{5} \implies \frac{8}{a+b} = -\frac{8}{5} \implies a+b = -5$.The roots of the required quadratic equation are $2a$ and $2b$.Sum of roots $= 2a + 2b = 2(a+b) = 2(-5) = -10$.Product of roots $= (2a)(2b) = 4ab = 4(4) = 16$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Therefore,$x^2 - (-10)x + 16 = 0$,which simplifies to $x^2 + 10x + 16 = 0$.

The harmonic mean of two numbers is $-\frac{8}{5}$ and their geometric mean is $2$. The quadratic equation whose roots are twice those numbers is

Similar Questions

Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x - 2 = 0$,where $\alpha > \beta$. If $a_n = \alpha^n - \beta^n$ for all $n \geq 1$,then what is the value of $\frac{a_{10} - 2a_8}{2a_9}$?

What is the geometric mean of the roots of the equation $x^2 - 18x + 9 = 0$?

If $\alpha$ and $\beta$ are the roots of $x^2 - ax + b = 0$ and if $\alpha^n + \beta^n = V_n$,then:

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$,then the roots of $ax^2-bx(x-1)+c(x-1)^2=0$ are

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then what is the value of $\alpha \beta^2 + \alpha^2 \beta + \alpha \beta$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module