Quadratic Equations
Tags: Engineering Entrance  |  JEE Main  |  CBSE Board  |  JEE Main & Advanced  |  Algebra
  « Back to Content



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.


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

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

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

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.

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.
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).


  Electricity and Effect of Current      Centre of Mass      Work Power & Energy      Magnetic Effects of Current      AIEEE 2009 Solutions and Analysis      English Core-XII Sample Test Paper      Function Theory      Basis of Inheritance      dasda      Business Studies -XII Sample Test Paper      Chemical Bonding      Group I A      Integration Theory      Organic Chemistry Basic Concepts      Applications of Derivatives      Informatics Practices (065) Sample Question Paper -III      Inorganic Chemistry Group VI A      Properties of Matter      Heat & Thermodynamics      Disha - Path to Success Series     

Free Sign Up!
Sponsored Ads

Preparing for JEE?

Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015

@ INR 5,443/-

For Quick Info