Roots of Equations | PRMO-2016 | Problem 8

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


quadratic equation


Check the Answer


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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Equations and roots | B.Stat Objective | TOMATO 71

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Equations and roots.

Equations and roots (B.Stat Objective problems)

The number of ordered pair of integers (x,y) satisfying the equation \(x^{2}+6x+y^{2}=4\) is

  • 2
  • 4
  • 6
  • 8

Key Concepts




Check the Answer

Answer: 4

B.Stat Objective Question 71

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

First hint

\(x^{2}+6x+y^{2}=4\) gives \((x+3)^{2}\)=13-\(y^{2}\)

Second Hint

for 1st value positive integer of x, we have 1st and 2nd value of y and for 2nd value negative integer of x we have 3rd and 4th value of y

Final Step

then number of ordered pair is 2+2=4.

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