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Consider a quartic equation

$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$

The value of $k$ so that given equation has three real and distinct solutions can be equal to - i think that for any four degree to have 3 roots, its $f'(x)$ should have two zero and by the same way its $f''(x)$ should have one zero which means that the determinant of $f''(x)$ should be zero but this is wrong since the correct answer to this question is $\frac{13}{2}$.

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    $\begingroup$ $f'(x)$ can have three zeros even if $f(x)$ has only three zeros. $\endgroup$ – Thomas Andrews Jul 23 '17 at 17:15
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    $\begingroup$ You know you need a repeated root, so if $f(x)$ has a repeated root, then $f(x)$ and $f'(x)$ have a common root. $\endgroup$ – Thomas Andrews Jul 23 '17 at 17:15
  • $\begingroup$ The above is true because (assuming $k$ is real) a real polynomial has roots that come in conjugate pairs, so if we have three real and distinct root, we must have 4 real roots ones, one repeated. $\endgroup$ – Zain Patel Jul 23 '17 at 17:17
  • $\begingroup$ I assume you meant "discriminant" not "determinant." $\endgroup$ – Thomas Andrews Jul 23 '17 at 17:18
  • $\begingroup$ @ZainPatel To what are you referring when you say "The above..." $\endgroup$ – Thomas Andrews Jul 23 '17 at 17:19
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A simpler approach: For a symmetric polynomial, if $x$ is a root, then so is $\frac{1}{x}$. So for there to be only three real roots, one of the roots must be $\pm 1$.

But that means either $13-2k=0$ or $13+2k=0$. Then you have to check that there are two other real roots in those cases.


This is related to the solution others have given by creating a polynomial of $x+\frac{1}{x}$ - any symmetric polynomial (that is, with coefficients reading the same forwards and back) can be written as a polynomial of $x+\frac1x$ times a power of $x$.

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Hint Observe that $0$ is not a root of this equation and since three real roots exist, therefore fourth must be real as well and there must be a repeated root.

Start with: \begin{align*} x^4 – kx^3 + 11x^2 – kx + 1 & = 0\\ x^2 – kx + 11 – \frac{k}{x} + \frac{1}{x^2} & = 0\\ \left(x+\frac{1}{x}\right)^2-k\left(x+\frac{1}{x}\right)+9 & =0\\ t^2-kt+9 & =0 && \left(\text{where } t=x+\frac{1}{x}\right) \end{align*} Now can you proceed from here to impose conditions for real roots?

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HINT: this equation is equivalant to $$x^2+\frac{1}{x^2}-k\left(x+\frac{1}{x}\right)+11=0$$ and set $$t=x+\frac{1}{x}$$

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Ignoring the specific form of your polynomial, a degree 4 polynomial can have three distinct real roots only if one of them is a double root (any complex roots would come in pairs). But $f(x)$ has a double root if and only if it has a common root with $f'(x)$. Hence you could perform Euklid's algorithm with $f(x)$ and $f'(x)$

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You already know a polynomial having one distinct root whose derivative has one root: $f(x) = x^2$. It is not automatic that the number of distinct roots decreases when you take the derivative of a polynomial -- the total number of roots (including multiplicities) decreases by one. So if it is one of the multiple roots that is removed, the number of distinct roots is unchanged.

A different way to go after $k$, which shows that $k = -13/2$ also works (and explains why $k = \pm 6$ does not).

A fourth degree polynomial has four roots. Non-real roots come in conjugate pairs, so if three roots are real, all four roots are real. If there are only three distinct real roots, one root is duplicated. Therefore, your polynomial factors as \begin{align*} p(x) = (x-a)^2(x-b)(x-c) \text{.} \end{align*} (We may take all coefficients of the "$x$"s in the factors to be $1$ because the given polynomial has leading coefficient $1$. This isn't essential, but it is a little convenient. If the given polynomial's leading coefficient were not $1$, we would merely multiply $p$ by this coefficient.) Multiplying out, we get \begin{align*} p(x) = x^4 &+ (-2a-b-c) x^3 + (a^2 + 2ab + 2ac + bc) x^2 \\ &+(-a^2 b - a^2 c - 2 a b c)x + a^2 b c \text{.} \end{align*} Comparing coefficients, we get \begin{align*} 1 &= 1 & [x^4] \\ -k &= -2a - b - c & [x^3] \\ 11 &= a^2 + 2ab + 2ac + bc & [x^2] \\ -k &= -a^2 b - a^2 c - 2 a b c & [x^1] \\ 1 &= a^2 b c & [x^0] \\ \end{align*}
We can solve this system. (Useful observations, $c[x^1]$ can eliminate its first RHS term with $[x^0]$, $b[x^1]$ eliminates the second, ans $a[x^1]$ eliminates the third. $a[x^2]$ eliminates $a^2b + a^2c$ with $[x^1]$ and with a numerical coefficient, eliminates $abc$ with it. And so on...) We eventually find solutions with $k = \pm 6$ or $k = \pm 13/2$, but when $k = \pm 6$, $b=c$, so there are only two distinct real roots and we must discard those potential solutions. Therefore, $k = \pm 13/2$. Then $a = \pm 1$, with the same sign as $k$ and $b,c = \pm \frac{1}{4}(9 \pm \sqrt{65})$, where the leading $\pm$ matches the sign of $k$ and the inner $\pm$ distinguishes $b$ from $c$. This gives all solutions and both satisfying values of $k$.

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