Let two numbers have an arithmetic mean of 9 | vedclass

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

Question

(B) Let the two numbers be $x_1$ and $x_2$.The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.The geometric mean

Vedclass · Accepted Answer

$x^2 - 18x + 16 = 0$

Vedclass · Answer

(B) Let the two numbers be $x_1$ and $x_2$.The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.The geometric mean is given by $\sqrt{x_1 x_2} = 4$,which implies $x_1 x_2 = 4^2 = 16$.$A$ quadratic equation with roots $x_1$ and $x_2$ is given by the formula: $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - 18x + 16 = 0$.

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

Similar Questions

If $a < 0$,then the inequality $ax^2 - 2x + 4 > 0$ has the solution represented by:

$I. 24x^2 + 38x + 15 = 0$
$II. 12y^2 + 28y + 15 = 0$

If the roots of the equation $12x^2 - mx + 5 = 0$ are in the ratio $2 : 3$,then $m =$

Solve the given two equations and select the correct option.
$I.$ $676 x^{2}-1=0$
$II.$ $y=\frac{1}{\sqrt[3]{13824}}$

If the roots of the equation $qx^2 + px + q = 0$,where $p$ and $q$ are real,are complex,then the roots of the equation $x^2 - 4qx + p^2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module