Question

find the quadratic equation whose zeroes are log 1000,log 0.01 0.1​

Answer 1

Answer:

x² -  5 x + 6

Step-by-step explanation:

Given :

Two zeroes of polynomial :

$\displaystyle \log1000 \quad , \quad \log_{0.1}0.01$

Rewrite as :

$\displaystyle \longrightarrow \log10^3 \quad , \quad \log_{1^{-1}}1^{-2}$

We have logarithm power rule if :

$\displaystyle \sf \longrightarrow \log m^n=n.\log m$

Using above formula we get :

$\displaystyle \longrightarrow \log10^3=3 \quad \text{and} \quad \log_{1^{-2}}1^{-2}=-2/-1\implies2$

Now sum of zeroes :

= > α + β = 3 + 2

= > α + β = 5

Product of zeroes :

= > α . β = 3 × ( 2 )

= > α . β =  6

We know for required polynomial is given as :

= > k ( x² - ( α + β ) x + α . β )

Putting value here we get :

= > x² -  5 x + 6

Therefore , we get required answer!