Question

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

• A. $x^2-18x-16=0$
• B. $x^2-18x+16=0$
• C. $x^2+18x-16=0$
• D. $x^2+18x+16=0$

Answer 1

The correct option is B

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

Explanation of the correct option:

Finding the quadratic equation:

Given the arithmetic mean as $9$ and the geometric mean as $4$.

Let $a$ and $b$ be the two numbers in arithmetic and geometric sequences

We know that the arithmetic mean is $\frac{\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{2}}$,

$\begin{array}{rcl}& \Rightarrow & \frac{a+b}{2}=9\\ & \Rightarrow & a+b=2\times 9\\ & =& 18\end{array}$

Also, The geometric mean is $\sqrt{\mathbf{a}\mathbf{b}}$,

$\begin{array}{rcl}& \Rightarrow & \sqrt{ab}=4\\ ab& =& 4^2\\ & =& 16\end{array}$

Using the obtained root the quadratic equation can be defined as $x^2-\left(a+b\right)x+ab=0$.

Now substitute $a+b$ as $18$ and $ab$ as $16$ in $x^2-\left(a+b\right)x+ab=0$.

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

Therefore, the equation is obtained as $x^2-18x+16=0$.

Hence, option (B) is the correct answer.