# Roots of Equations | PRMO-2016 | Problem 8

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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

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........

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

# Equations and roots | B.Stat Objective | TOMATO 71

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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

Roots

Integers

Equations

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.