let the quadratic eq ax^2 +bx + c be such that a,b,c are distinct and each of a,b,c belong to {1,2,3.....n} such that x+1 divides ax^2+ bx +c.If such quad.polynomials are < 99,then max(n)=?

a)14 b)15 c)16 d)18 e)none of these

The number of non-negative integer solutions of a + b = c ---> ^{c – 1}C_{1} = c - 1. Therefore, number of solutions when c = 2, 3, 4, .., 15 = 1, 2, 3, 4, ... 14 = 105. Out of these we have to remove solutions where a and b are not distinct. it happens when c is even. The number of cases = 7. Therefore, total solutions = 105 - 7 = 98 and answer is b.

if y is an integer such that y>=4 and x=y*y -2y,then largest number that divides x*x-8x=?

If you substitute x=y^{2}-2y into x^{2}-8x you will obtain,

x^{2}-8x = (y-4)(y-2)(y)(y+2)

__If y is odd__:

then x^{2}-8x is divisible by **3**

__If y is even__:

then it has to be divisible by 2 (4!) = **48**

For the equation 4x^{2} - 2xy - 8x + y + 9 = 0, find the number of positive integral solutions.

Hi Tina,

Plese find the solution

We want to find number of positive integral solutions for (x,y) 4x^{2} - 2xy - 8x + y + 9 = 0

=>4x^{2 }- 8x + 9 = (2x - 1)y

=>y = (4x^{2} - 8x + 9)/(2x - 1) = (2x -3) + 6/(2x-1).

Only two values satisfy. **(1,5), (2,3)**

Hello sir, please explain

Hello Kinshuk !

a^19 + 17a - 19 = 0

a^19 = 19 -17a

similarly , b^19 = 19 - 17b ,

c^19 = 19 - 17c .....

So a^19 + b^19 + c^19 +....... = 19 - 17a + 19 - 17b + 19 - 17c +......

= 19 × 19 - 17 ( a+ b + c +....)

Sum of roots => 0

19 × 19 - 17 × 0 = 361.

For how many integer values of *x* is an integer?

Plz solve this que.

Hi Gaurav ! PFA the Solution

22nd: 3s+5t=750. First solution will be, s=0, t=150. Now, value of t will go down by 3 (which is coefficent of s) for every next solution. Next solution is s=5, t=147, then next will be, s=10, y=144 and so on. Value of t will vary from 150 to 0, with common difference of 3. Hence number of solutions will be {(150-0)/3} +1=51.

If you want to find out natural solutions, then 51-2=49. We subtract 2 because of (0,150), and (250,0).

The number of quadratic equation with leading coefficient 1 which are unchanged by squaring their roots is?

Hello Yamin ,

a and b are roots

So , a + b = a² + b²

and ab = a²b²

=> ab(ab - 1)

i) a = 0

=> b = 0 or 1

ii) b = 0

=> a = 0 or 1

iii) ab = 1

a + 1/a = a² + 1/a²

=> a⁴ - a³- a + 1 = 0

=> (a³ - 1)(a - 1) = 0

=> a = 1 or w or w²

=> b = 1 or w² or w

So, equation are x² = 0

x² - x= 0

x² - 2x + 1 = 0 and x² + x + 1 = 0

4 such equations.

If the roots of x^2 - 2ax + a^2 + a -3 are real and less than 3 then

A. a is less than two

B. a lies between 2 and 3

C. a is greater than 4

Hello Yamin,

put x = 0 in the expression x² - 2ax + a² + a - 3

it reduces to

a² + a - 3 = 0

a = -1/2 + (√13)/2 and -1/2 + √13/2

Here roots are less than 2 hence option (B) and (C) got eliminated.

Hence , Option (A ) is the correct answer .

Alternate Approach :

If eq x^3 - ax^2 + bx - a =0 has 3 real roots the what is true?

A. a=1

B. b=1

C. b not equal to 1

x³ - ax² + bx -a = 0

x² ( x - a) + b ( x-a) = 0

if , b = 1 , then the equation reduces to

x²( x - a) + (x-a) = 0

(x-a)(x² + 1) = 0

it has one real root a and two imaginary roots . hence b ≠ 1 .

Option C.

3x^2 + px + 3 has p greater tha 0. One root is square of other. Then what is the value of p?

Hello Yamin,

Sum of roots = a + a² = -p/3 ..... (1)

Product of the roots = a³ = 3/3 = 1 ... (2)

a³ -1 = 0

(a-1)( a² + a + 1 ) = 0

(a-1) = 0 or ( a² + a + 1 ) = 0

a² + a = - 1

From equation (1)

-1 = -p/3

p = 3 🙂

In what positive base b does 4×12=103 hold?

Hello Yamin,

4 × 12 = 103

4× ( 1•b + 2) = 1•b² + 0•b + 3

4b + 8 = b²+ 3

b² - 4b - 5 = 0

b² - 5b + b -5 = 0

b ( b -5) + 1 ( b -5) = 0

( b +1) ( b -5) = 0

b = -1 or b = 5

b = -1 is inadmissible hence b = 5 .