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Quadratic Equations
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Illustrations

Medium

Q-1: Prove that  has no integral solution

Solution: Let  then the equation is

As u, v are integers, >1402

But 372 = 1369, 382 = 1444

Minimum possible value of v=38.

Also

v must be even.

Let v=38+2k where k N





Hence D=0

i.e.  which is incorrect.

u, v cannot have positive integral solutions.

So, x, y cannot have integral solutions.
Q-2: If a, b are the roots of  and also of  and if  are the roots of  then prove that n must be even integer
Solution:

Sincea, b are the roots of





This is true only if n is an even integer.
Q-3: Show that the equation
 has no imaginary roots,
Where A, B, C---------K, a, b, c------k and l are real.

Solution: Assume a+ib is an imaginary root of the given equation then conjugate of this root a-ib is also root of this equation.

Putting x = a+ib and x = a-ib in the given equation then,

The expression in bracket 0

2ib = 0 b = 0 (because i 0)

Hence all roots of the given equation are real.
Q-4: Solve in
Solution:

The A.M of x+3, x-1 is  , i.e. (x+1)
Put x+1 = y then x+3 = y+2, x-1 = y-2

The equation becomes



The corresponding equation =0 has roots 1, -1.
The sign scheme of  y R is as follows.

Fig (13)
y2-1, 0 holds if y -1 or y 1
Now y -1 x+1 -1 or x -2
y 1 x+1 1 or x 0
Hence x -2 or x 0
Therefore the solution set = .
Q-5: let a, b, c be real, if  has two real roots a and b where a< -1 and
b >1 then show that
Solution:



Fig (14)      
Q-6: If a<b<c<d prove that the equation (x-a) (x-c) + k (x-b) (x-d) = 0 has real roots for all k R.
Solution:



Equation (1) will have real roots if,


Now the discriminant of the equation corresponding (2) is,





Because a<b<c<d

Also (b-d)2>0 so (2) is true for all k R.
Hence given equation have real roots.
 
Hard
Q-1: If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means between the same two numbers, prove that q cannot lie between

Solution: Let the numbers be x, y. Let the n AMs between them be A1, A2, -------- An

Then y = (n+2) th term = x + (n+1) d          (d being the common difference)



Equation (1) Þ y = (n+1) p-nx putting this in (2).



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