Find the number of values of k for which the expression

$$y= (x^2 - x(k-2) + k^2)(x^2 + kx + (2k-1))$$

is a perfect square .

I tried by assuming the product as $$ t^2 $$ And then assuming two separate equations as t Is this the right approach.

How i can achieve this? Thanks in advance.


Another approach is to multiply out the expression to get


If this is a perfect square, it will have the form


for some $a$ and $b$. Equating coefficients, we see that $a=1$ and

$$\left\{\begin{align*} 2b+1&=4k-1\\ 2b&=k^3-2k^2+5k-2\\ b^2&=k^2(2k-1)\;. \end{align*}\right.$$

The first two of these equations are easy to solve, and the results can then be tested against the third equation.

  • $\begingroup$ I think this rather straight forward and more nicer than mine. +1 $\endgroup$ – user21436 Mar 17 '12 at 7:05
  • $\begingroup$ @Kannappan: Yours works, though, so I gave it a +1. $\endgroup$ – Brian M. Scott Mar 17 '12 at 7:31

You may want to look at the following cases:

  1. Each of the quadratic factor of $y$ is a square.

  2. Each of the quadratic factor has exactly the same linear factors.

  3. Each of the quadratic factor is the same irreducible.

Actually, $(2)$ and $(3)$ tell you that the quadratic factors are the same upto multiplicative constants.

I'll work the details if you still cannot break ice from here.

  • $(1)$ Each of the quadratic factor of $y$ is a square.

Then, you must have that discriminant must equal $0$ for each quadratic factor. $$(k-2)^2-4k^2=0 ~~~~\mbox{and}~~~~ k^2-4(2k-1)=0 $$ Now, solve this equation to see that $k=\dfrac {16}{28}$ but this does not satisy either equation. So, they don't have common roots. (Alternatively, solve both quadratic equations using the quadratic formula, you know and check that no root of one is a root for the other.)

  • $(2), (3)$ The two equations are one and the same upto some multiplicative constants.

In this case, the coefficients are proportional to each other.


What does this get you? Using the fact that $k^2-2k+1=0$ (How did I get this?), get that $$\boxed{k=1}$$ is the only solution.

Hope this helps. I also suggest reading Brian's approach.

  • $\begingroup$ sorry.i am not able to get anything from this.Now i am thinking i am very weak in mathematics. $\endgroup$ – vikiiii Mar 17 '12 at 7:00
  • $\begingroup$ can you please explain your answer? $\endgroup$ – vikiiii Mar 20 '12 at 13:08
  • $\begingroup$ @vikiiii I have explained as you sought... $\endgroup$ – user21436 Mar 20 '12 at 13:57

$$y= (x^2 - x(k-2) + k^2)(x^2 + kx + (2k-1))$$

$$y= (x-a)(x-b)(x-c)(x-d)$$ for some real numbers, a,b,c,d

for the equation to represent a perfect square
either a=b and c=d .....................(1)
or a=c and b=d ..........................(2)
or a=d and b=c ..........................(3)

Condition 1:
The discriminant of both the equations must be 0, simultaneously. when we equate the discriminant of the first equation to 0, we get , k= -2, k=2/3
for second equation we get irrational values of k. hence Condition 1 fails to be satisfied.

Condition 2 and Condition 3:
The two quadratics must have a root in common, for these 2 condition to be satisfied.
let that common root be $x=t$,

$$ t^2 - t(k-2) + k^2= t^2 + kt + (2k-1) = 0$$

$$k^2-2k+1 = 2t(k-1)$$


this gives $k=1$,

  • $\begingroup$ Please read my answer to see where your answer is faulty. (Andre's answer is also good.) $\endgroup$ – user21436 Mar 20 '12 at 13:58

A polynomial $P(x)$ is the square of another polynomial iff the (complex) roots of $P(x)$ come in pairs. Let the roots of our product be $\alpha$, $\alpha$ and $\beta$, $\beta$ (maybe $\alpha=\beta$).

Perhaps each quadratic we see has the roots $\alpha$ and $\beta$. Since the sum of the roots is the negative of the coefficient of $x$, we must then have $k-2=-k$, that is, $k=1$. Does that work? Sure, because then $k^2=2k-1$, so the constant terms also match.

Or else the roots of the two quadratics are respectively $\alpha$, $\alpha$, and $\beta$, $\beta$. This happens iff each discriminant is $0$. So we want $$(k-2)^2-4k^2=0\quad\text{and}\quad k^2-8k+4=0.\qquad\qquad(\ast)$$ But $(\ast)$ forces $(-3k^2-4k+4)+3(k^2-8k+4)=0$. We get a linear equation for $k$, with integer coefficients. That would make $k$ rational. But the roots of $k^2-8k+4=0$ are not rational.

Alternately, we could find the roots of the two quadratics in $(\ast)$, by formula or by calculator, and see that the roots of one are not roots of the other.

Thus the only value of $k$ that makes our product a perfect square is $k=1$.


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