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AMC 8 Math Olympiad

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

Quadratic equation

Roots

Check the Answer


Answer: (b)

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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AMC 8 Math Olympiad

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

Check the Answer


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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AMC 8 Math Olympiad

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

Check the Answer


Answer: 10

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

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