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Algebra Level 4

The number of quadratic equations with real roots which remain unchanged even after squaring their roots is:

The answer is 3.

2 solutions

Jonathan Gray
Jan 11, 2015

If $\alpha, \beta$ are as desired, then $(x-\alpha)(x-\beta)=(x-\alpha^2)(x-\beta^2)$ Thus $\alpha^2+\beta^2=\alpha+\beta$ and $\alpha^2\beta^2=\alpha\beta$ Hence $\alpha,\beta=0,1$ . This yields three distinct quadratics.

Just in case anyone is wondering, the quadratic equations will be $x^2=0~,~(x-1)^2=0\textrm{ and } x(x-1)=0$ .

Prasun Biswas - 6 years, 5 months ago

the answer should be 4 let the roots be a and b then we know that a+b=a^2+b^2 and ab=(ab)^2 ab(ab-1)=0 so one is a=0 so b=0 or1 same is when we take b=0 so one is with roots 0,0 and 0,1 but when ab=1 so (a+b)^2-2ab=a+b so putting a+b=x x^2-x-2 x=2or -1 by using x^2-sx+p=0 so eq are x^2-2x+1 and 2nd is x^2+x+1

so equation with roots (0,0),(0,1),(1,1)and (w,w^2)
will not change if squared hence no. of equations is 4

madhav gupta - 6 years, 4 months ago

I also agree with u

Ashutosh Kaul - 6 years, 3 months ago

I said real roots

Kushal Patankar - 6 years, 2 months ago

Real roots

علاء رمضان - 4 years, 3 months ago

Jonathan Hocker
Nov 3, 2016

I think you should add monic to the description as you can multiply the three types of quadratics by a constant to form an infinite number of quadratic equations with the same roots.

the first answer I came up with was infinite, but I couldn't enter it as an answer, as an integer is required

Iris B - 4 years, 5 months ago

Require some Smartness

The answer is 3.

2 solutions

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