Question

Let the two numbers have an arithmetic mean of 9 and a geometric mean of 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

Hint: In this question, we are given the arithmetic mean and the geometric mean of the quadratic equation and we have to find the quadratic equation. First, we use the formulas of AM and GM to find the sum and the product of roots. By using the roots and putting them in the standard quadratic equation, we get the desirable quadratic equation.

Formula used:
$AM=\dfrac{a+b}{2}$
And GM is $ab={{c}^{2}}$
And ${{x}^{2}}$- ( sum of roots a and b )$x$+ ( product of roots a and b ) = 0

Complete step by step Solution:
Let $\alpha $ and $\beta $ be two numbers whose arithmetic mean is 9 and geometric mean is 4
We have to find the quadratic equation with the above roots.
As arithmetic mean is 9
Then $\dfrac{\alpha +\beta }{2}=9$
That is $\alpha +\beta =18$
And geometric mean is 4
We know GM is $ab={{c}^{2}}$
That is $\alpha \beta ={{(4)}^{2}}$
Then $\alpha \beta =16$
We know any quadratic equation whose roots are a and b can be formed by writing as
${{x}^{2}}$- ( sum of roots a and b )$x$+ ( product of roots a and b ) = 0
Quadratic equations whose roots are a and b are given as
${{x}^{2}}-(a+b)x+ab=0$
Then the quadratic equation whose roots are $\alpha $and $\beta $is
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0$
Now By putting the values of $\alpha +\beta $and $\alpha \beta $in the above equation, we get
${{x}^{2}}-18x+16=0$

Therefore, the correct option is (B).

Note: To solve these types of questions, students must know about arithmetic and geometric mean. Arithmetic mean is simply the sum of numbers divided by total numbers. And the geometric mean is the mean which tells us about the tendency of the set of numbers by finding the product of their values.