Equations and Expre...

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# Equations and Expressions

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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

Since, x+1 satisfies the equation ax^2+ bx+ c=0. Therefore, a(-1)^2+b(-1)+c=0. So, a - b+ c=0.
Now, a + c=b. Minimum value of b is 3. Number of quadratic equations will exceed 99 when b=12. So, the answer is none of these.

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Find the condition for the equation ax^2 + bx + c = 0 for one root is n times the other

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X=2+2^2/3+2^1/3, then the value of x^3-6x^2+6x?

Hello Tarishi

(x - 2) = 2⅔ + 2⅓

Cubing both sides

(x - 2)³ =( 2⅔ + 2⅓ )³

= (2⅔)³ + ( 2⅓)³ + 3•2⅔• 2⅓( 2⅔ + 2⅓ )

[( a + b)³ = a³ + b³ + 3ab ( a + b )]

6 + 3 × 2(2⅔ + 2⅓) = 6 + 6(x - 2)

x³ - 8 - 6x² + 12x = 6 + 6x - 12

x³ - 6x²  + 6x = 6 - 12 + 8 = 2.

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Solve the equation:x^4-2x^3+4x^2+6x-21=0 if two of its roots are equal in magnitude but opposite in sign .

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If the constant term of (2x-1/x)^n is -160 ! Find n

Hello Tarishi ,

General term of (2x – 1/x) n   = nCr (2x)r (-1/x)(n-r)

= nCr (2r) (-1)(n-r) xr(1/x)(n-r)

Now ,
for constant term , power of x = 0

r = n – r

n = 2r

Constant Term = 2rCr2r (-1)r= - 160

r = 3 satisfies , Hence n = 2r = 6 .

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Find a four digit perfect square such that its first two digits are the same and its last two digits are the same.

Hello Samyak ,

Let the four digit number be aabb

1100a + 11b = x²

Clearly , x is a multiple of 11

so , 1100a + 11b = (11y)²

100a+ b = 11y²

a + b mod 11 = 0

b = 11-a

100a +11 -a = 11y²

99a +11 = 11y²

9a +1 = y²
Only possibility , a = 7,

So , b = 4.

aabb = 7744

Alternate Approach :

We know 122 ends with __44, i.e last 2 digits repeating

So the square must be (50k - 12)2 or (50k + 12)2

122 = 144
382 = 1444
622 = 3844
882 = 7744

Hence , 7744.

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p² + q² = ( p+ q)² - 2pq

We know , sum of the roots = -  ( Coefficient of x) / ( Coefficient of x²)

So , p + q = ( a -2)

and the product of the roots = ( constant term )/ ( coefficient of x² )

=> pq = - ( a +1)

p² + q² = (a -2)² + 2( a + 1)

= a² + 4 - 4a + 2a + 2

= a² -2a + 6

= ( a -1)² + 5 .

minimum value of ( a -1)² is zero. Hence the minimum possible value of p² + q² is 5.

Option (D)

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In the expansion of (x - √x - 7)^7, what is the coefficient of x^(5/2)?

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Hello sir , Please provide the solution to this problem

open all the brackets with + sign and solve it will give you x<=36

open all the brackets with -ve sign and solve it will give you x>=(-64)

total integer values =36+64 +1              1 is for 0

=101 (answr)

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Lord of Mordor distributed some magical rings among his Nazguls. The first Nazgul received 100 rings and one tenth of the remaining rings. The second Nazgul received 200 rings and one tenth of the remaining rings. The third Nazgul received 300 rings and one tenth of the remaining rings... and so on. After distributing the rings in this way, the Lord found out that he had given equal number of rings to all the Nazguls. How many Nazguls were there?

A. 6

B.9

C. 12

D. 13

Let the total rings be n
1st  Nazgul recieved => 100 + (n - 100)/10 rings.

2nd Nazgul recieved=> 200 + 1/10(n - 300 - (n- 100)/10) rings .

since these both are equal, equate it.

n = 8100

so 1st Nazgul recieved = 100 + 8000/10 = 900

Since all person received same amount Hence , 8100/900 = 9 Nazguls

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A tells B , when I was 3/5th of your present age you were 5/7th of my present age . Present age of A is 28 years . B's Present age is ?

Hello Richa,

Let the present age of B be x years.

So ,when A was 3/5 of x. That time B was 5/7 of 28 = 20.

Now ,

28 - 3/5 x = x -20

{ Difference should be constant }

x = 30 .

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Sir,please provide the solution for both questions.

Solution to the 27th Question :

-2 , -1 , 1 and 2 are the roots of the equation so

f(x)=k•(x+2)(x+1)(x-1)(x-2).

f(p)=k(p+2)(p+1)(p)(p-1)(p-2)/p

=>  f(p) × p is a product of 5 consecutive numbers.

So,must be divisible by 5.

Now p, p+1, p+2 are 3 consecutive numbers,so should be divisible by 3,will say the same for p, p-1, p-2. So, divisible by 3² i.e 9.

p-1, p+1 are consecutive even numbers,must be divisible by 8(Because 2 × 2²).

Hence , divisible by 5, 9 and 8

Largest Integer =5 × 8 × 9 = 360.

Option (4)

28th

x² + 2y² + 4z² + 3 ( xy + yz + zx) = 16 ... ( 1)

y² + 2z² + 4x² + 3 ( xy + yz + zx) = 16 ... (2)

z² + 2x² + 4y² + 3 ( xy + yz + zx) = 16 ... (3)

On adding (1) , (2) and (3)

7 ( x² + y² + z²) + 9 ( xy + yz + zx) = 48

x² + y² + z² =3

xy+yz+zx=3

x=y=z=+1 or-1

so only 2 solutions are possible

hence option 2

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for a cubic polynomial ax3 + bx2+cx+d=0

three roots can be described as

p +q+r=-b/a

pq+qr+pr= c/a

pqr=-d/a

so in given equation

p+q+r=0.................(1)

pq+qr+pr=a.................(2)

squaring (1)

p2 +q2+ r2  +2(pq+qr+rp)= 0

p2 +q2+ r2  =-2(pq+pr+qr)

from (2)

p2 +q2+ r2  =-2a

a=-(p2 +q2+ r2  )/2

since roots are integers and squaring them will always give a positive value

hence a will always be negative integer

so