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

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 \(\frac{1+b+ab}{a}\)

- $200$
- $240$
- $300$

Algebra

quadratic equation

Roots

But try the problem first...

Answer:$240$

Source

Suggested Reading

PRMO-2016, Problem 8

Pre College Mathematics

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

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

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