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

quadratic equation

Roots

Check the Answer


But try the problem first…

Answer:$240$

Source
Suggested Reading

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

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

Subscribe to Cheenta at Youtube