10-seconds challenge-9

Algebra Level 2

If the minimum value of $x^{2}+x+1$ is $\frac{3}{4}$ . Find the maximum value of $\frac{x^{2}+x+4}{x^{2}+x+1}.$

2 solutions

Yash Dev Lamba
Mar 10, 2016

We can write $\frac{x^{2}+x+4}{x^{2}+x+1}=\frac{x^{2}+x+1+3}{x^{2}+x+1}=1+\frac{3}{x^{2}+x+1}$

For $\frac{x^{2}+x+4}{x^{2}+x+1}$ to be maximum $\implies$ $x^{2}+x+1$ to be minimum.

So maximum value of $\frac{x^{2}+x+4}{x^{2}+x+1}= 1+\frac{3}{\frac{3}{4}}=1+4=\boxed{5}$

This is 10 sec method.

You made it much more easier by giving the minimum value of the quadratic (one can find this on his own).

Akshat Sharda - 5 years, 3 months ago

Actually it is given to make it 10 sec challenge otherwise it would be 20 sec challenge

Yash Dev Lamba - 5 years, 3 months ago

Okay! I can understand it.

Akshat Sharda - 5 years, 3 months ago

Garvit Gosain
Apr 2, 2016

x x+x=3\4-1\1=x x+x=-1\4

x x+x+4 =-1\4+4\1. (By substituting the value of x x+x ) The value of x*x+x+1=3\4

15\4 ÷ 3\4 =5