IIT JEE , AIEEE Forum - Mathematics , Algebra

abhishekneha

the number of quadratic equations with real roots and remaining unchanged even after squaring their roots is............??

akhil_o

x²+bx+c=0

let roots be A,B

by condition

c²=c as product of roots = c, product of squares=c²

^{So we have} c=1

A²+B²=b²-2c

hence -b=b²-2c

or b²+b=2

b=-1 ir b=2

hence we have 2 quadratic eqns

x²-x+1=0

and

x²-2x+1=0

Correction Made..thanks to Konichiwa2x

only the second eqn has real roots
hence we have only one solution
x^2-2x+1=0

konichiwa2x

Just one.

akhil_o

Do we consider equations with unreal roots also?

one has real roots and the other unreal!

konichiwa2x

Here is a more general method.

Let the quadratic be

and let the roots be

.

We have,

and

.

From above equation,

or

.

Note that if one of the roots is 0(say

), then the other root(

) would become complex. Hence, we can conclude that the roots are recicprocals of one another.

Therefore,

So our equation simplifies to

with condition

(since it is given that roots are real).

Now,

.

=>

The above is a quadratic in b. Solving we get,

or

But if

, the discriminant would be less than 0 leading to complex roots.

So there is only one quadratic equation:

which simplifies to

or

.

konichiwa2x

no akhil, the question states that the quadratic has real roots. So dont consider the equation with complex roots.