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## Quadratic equation | ISI-B.stat | Objective Problem 240

Try this beautiful problem based on Quadratic equation, useful for ISI B.Stat Entrance.

## Quadratic equation | ISI B.Stat Entrance | Problem 240

The equations $x^2 + x + a = 0$ and $x^2 + ax + 1 = 0$

• (a) cannot have a common real root for any value of a
• (b) have a common real root for exactly one value of a
• (c) have a common root for exactly two values of a
• (d) have a common root for exactly three values of a.

### Key Concepts

Algebra

Roots

TOMATO, Problem 240

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Let the equations have a common root $α$.Therefore $α$ must satisfy two given equations…….

Therefore,

Now, $α^2 + α + a = 0$……………….(1)
And, $α^2 + aα + 1 = 0$…………………..(2)

Can you find out the value of $a$?

Can you now finish the problem ……….

Therefore,

Using cross-multiplication betwwen (1) & (2) we will get…….

$\frac{α^2}{(1 – a^2)} =\frac{ α}{(a – 1)} = \frac{1}{(a – 1)}$
$\Rightarrow {α}^2 = \frac{(1 – a^2)}{(a – 1) }=- (a + 1)$ & $α=\frac{(a-1)}{(a-1)}=1$

Now ${α}^2=(α)^2$
$\Rightarrow -(a+1)=1$
$\Rightarrow a = -2$

Therefore (b) is the correct answer….

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## Integer Problem |ISI-B.stat | Objective Problem 156

Try this beautiful problem Based on Integer, useful for ISI B.Stat Entrance.

## Integer | ISI B.Stat Entrance | Problem 156

Let n be any integer. Then $n(n + 1)(2n + 1)$

• (a) is a perfect square
• (b) is an odd number
• (c) is an integral multiple of 6
• (d) does not necessarily have any foregoing properties.

### Key Concepts

Integer

Perfect square numbers

Odd number

Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

## Try with Hints

$n(n + 1)$ is divisible by $2$ as they are consecutive integers.

If $n\not\equiv 0$ (mod 3) then there arise two casess……..
Case 1,,

Let $n \equiv 1$ (mod 3)
Then $2n + 1$ is divisible by 3.

Let $n \equiv2$ (mod 3)
Then$n + 1$ is divisible by $3$

Can you now finish the problem ……….

Now, if $n$ is divisible by $3$, then we can say that $n(n + 1)(2n + 1)$ is always
divisible by $2*3 = 6$

Therefore option (c) is the correct

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## Integer | ISI-B.stat Entrance(Objective from TOMATO) | Problem 72

Try this beautiful problem Based on Integer useful for ISI B.Stat Entrance.

## Integer| ISI B.Stat Entrance | Problem 72

The number of integer (positive ,negative or zero)solutions of $xy-6(x+y)=0$ with $x\leq y$ is

• 5
• 10
• 12
• 9

### Key Concepts

Integer

Algebra

Divisor

TOMATO, Problem 72

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Factorize the given equation

Can you now finish the problem ……….

Find the divisor

can you finish the problem……..

Given equation is $xy-6(x+y)=0$

$\Rightarrow xy-6x-6y=0$

$\Rightarrow xy-6x-6y+36=36$

$\Rightarrow (x-6)(y-6)=3^2 \times 2^2$

Now the numbers of factpr of $36=9$ i.e $\{1,2,3,4,6,9,12,18,36\}$

Thus we may say that 36 has 9 positive divisors, and 9 negative. and x=0 and y=0 is also a solution

the given condition $x\leq y$ ,so there are 10 non-negetive solution