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Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on quadratic equation from PRMO 2016. You may use sequential hints to solve the problem.

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


Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

Try with Hints


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 ……….

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

Can you finish the problem……..

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

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