IIT JEE , AIEEE Forum - Mathematics , Algebra

Jalaj

the number of quadratic equation/s which are unchanged by squaring their roots is?

elessar_iitkgp

The only numbers that remain unchanged on squaring are 0 and 1.
So the equations are
x²=0
(x-1)²=0
x(x-1) = 0

The answer seems to be three such equations

I had assumed real roots. But if complex roots are to be considered, there are four such equations as has been said in the posts below.

neeraj_agarwal_1990

I agree...

Oldisgold

elessar_iitkgp is right. There are 2 more extreme possibilities as well, as below

The following are the total possibilities (theoretically).

1. (X - 0) * (X - 0)

2. (X - 0) * (X - 1)

3. (X - 1) * (X - 0) [same as 2 above. But enumerated for completeness)

4. (X - 1) * (X - 1)

5. (X - 0) * (X -

) [ one root being 0 and the other being infinity. one may argue

whether infinity can have a square or any exponent.The

moment v think of infinity as a concrete no, then it can. But

if it is an abstract entity, it cant. But then, it should not

have been used in calculations at all, if abstract and not

concrete.Hence, for the sake of completeness, the following

may need to be included in a maths-sense.

Am I being Polemical ? Perhaps, But then what is science

without its share of diversity of opinion.anyway. Remember

Bayesian vs classicists schools wrt Prob theory ? ]

]

6. (X -

) * (X - 0)

7. (X -

) * (X -

)

mrityunjay

ans is 4.

               x²=0
                (x-1)²=0
                x(x-1)=0
                x² + x +1 =0

Jalaj

yups ans is 4

ambareesh13

so the question is already solved ......the last one are roots of unity(omega and omega square) ...so the ans is 4

Betu

how the roots of x²+ x +1 =0 are same when squared.....pl explain

neeraj_agarwal_1990

roots of x^2 +x+1 =0 are omega and omega ^2...
when they are squared...we get omega ^2 and omega ^4 ..
and omega^3=1....so roots are omega ^2 and omega...which are same as before..

raghavnegi

ans 4 eqs

spideyunlimited

ans is 4.

let original eqn be
x^2 - x(a + b) +ab = 0

squared roots eqn will be
x^2 - x(a^2 + b^2) + a^2.b^2 = 0

the eqn will be unchanged iff ( if and only if)
1) a^2 + b^2 = a + b
2) ab = a^2. b^2

we get an eqn by substituting b^2 = ab / a^2

final eqn is
a^4 - a^3 - a^2.b + ab = 0
degree of eqn is 4 therefore 4 solutions hence four quadratic eqns are possible.

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