 Try this beautiful problem from Algebra based on roots of equations.

## Roots of Equations | PRMO | Problem 8

Suppose that $a$ and $b$ are real numbers such that $ab \neq 1$ and the equations $120 a^2 -120a+1=0$ and $b^2-120b+120=0$ hold. Find the value of $\frac{1+b+ab}{a}$

• $200$
• $240$
• $300$

### Key Concepts

Algebra

Roots

But try the problem first…

Answer:$240$

Source

PRMO-2016, Problem 8

Pre College Mathematics

## Try with Hints

First hint

The given equations are $120 a^2 -120a+1=0$ and $b^2-120b+120=0$.we have to find out the values of $a$ and $b$….

Let $x,y$ be the roots of the equation $120 a^2 -120a+1=0$then $\frac{1}{x},\frac{1}{y}$ be the roots of the equations of $b^2-120b+120=0$.can you find out the value of $a$ & $b$

Can you now finish the problem ……….

Second Hint

From two equations after sim[lificatiopn we get…$a=x$ and $b=\frac{1}{y}$ (as $ab \neq 1)$

Can you finish the problem……..

Final Step

$\frac{1+b+ab}{a}$=$\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}$=$\frac{(x+y)+1}{xy}=240$