# Quadratic Equation - 5

__Medium__

Q-1: Prove that has no integral solution

Solution: Let then the equation is ** **

As u, v are integers, >1402

But 37^{2} = 1369, 38^{2} = 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)

y^{2}-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) Ãž y = (n+1) p-nx putting this in (2).