Question

find two numbers whose A.M. is 50 and GM is 40
​

Answer 1

$\large\underline{\text{Understanding the Problem}}$

Let us assume the two numbers are $a$ and $b$.

As per the definition of A.M.,

$\dfrac{a+b}{2}=50$

$\iff\underline{a+b=100}$

As per the definition of G.M.,

$\sqrt{ab}=40$

$\iff\underline{ab=1600}$

$\large\underline{\text{Solution}}$

Now, we know that $a+b=100$ and $ab=1600$.

The polynomial which the zeros are two numbers is $x^{2}-100x+1600=0$.

Let $10t=x$.

$100t^{2}-1000t+1600=0$

$\iff t^{2}-10t+16=0$

$\iff(t-2)(t-8)=0$

$\iff t=2\text{ or }t=8$

According to the substitution,

$\iff10t=20\text{ or }10t=80$

$\iff x=20\text{ or }x=80$

$\large\underline{\text{Answer}}$

Hence, the two numbers are 20 and 80.

Answer 2

Question

➽find two numbers whose A.M. is 50 and GM is 40

$\boxed{ \gray{ \huge \sf \: a.m}}$

➽ a+b/2 = 50

➽ a+b = 50×2

➽ a+b= 100

$\boxed{ \gray{ \sf \huge \:g.m}}$

➨√ab = 40

➨ab = 40 × 40

➨ ab = 1600

The polynomial with the zero two number is

x²-100x + 1600 = 0

➽ let 10t = x

➽ 100t² - 1000t -1600 = 0

➽ t² - 10t + 16 = 0

➽ ( t - 2 )( t - 8 ) = 0

✠ t = 2 or t = 8

So,

➽ 10t = 20 or 10t = 80

➽ x =20 or 10t = 80

Hence,

★ 20 and 80 Answer.