Polynomials (roots) (1 Viewer)

senarillosk · Nov 30, 2015

Two roots of x³ + mx² + 15x -7 = 0 are equal and rational. Find m.

Would appreciate it if someone explained this to me. Thank you.

InteGrand · Nov 30, 2015

senarillosk said:
Two roots of x³ + mx² + 15x -7 = 0 are equal and rational. Find m.

Would appreciate it if someone explained this to me. Thank you.

$Let the double root (which is rational) be $\alpha$. By the rational roots theorem, since the polynomial is monic (leading coefficient is 1) $\alpha$ must be a factor of the constant term, so the test values for $\alpha$ are $\pm 1,\pm 7$ (as these are the factors of the constant term $-7$).$

$Note that as $\alpha$ is a double root, it must also be a root of the derivative, namely the polynomial $P^\prime (x) = 3x^2+2mx+15$.$

$So we need to see which of the test-values for $\alpha$ will satisfy \textsl{simultaneously} the given polynomial equation$

$x^3 + mx^2+15x-7=0 \quad (1)$

$and the equation$

$$3x^2+2mx+15=0 \quad (2)$.$

$Substituting $x=1$ into (2) gives us $3+2m+15 = 0\Rightarrow m = -9$.$

$Then substituting $x=1$ into the L.H.S. of (1) with $m=-9$ gives$

$\begin{align*}1-9+15-7 &= 0 \end{align*}$

$so Equation (1) is also satisfied, so $m=-9$ with the double root being $1$ works.$

$Substituting $x=-1$ into (2) gives us $3-2m+15 = 0\Rightarrow m = 9$.$

$Then substituting $x=-1$ into the L.H.S. of (1) with $m=9$ gives$

$\begin{align*}-1+9-15-7 &= -14 \neq 0 \end{align*}$

$so Equation (1) is not satisfied, so a double root of $\alpha =-1$ does not work.$

$Substituting $x=7$ into (2) gives us $3\times 49+14m+15 = 0\Rightarrow m = -\frac{81}{7}$.$

$Then substituting $x=7$ into the L.H.S. of (1) with $m=-\frac{81}{7}$ gives$

$\begin{align*}7^3+7^2 \times \left( -\frac{81}{7} \right) +15\times 7-7 &= -126 \neq 0 \end{align*}$

$so Equation (1) is not satisfied, so a double root of $\alpha =7$ does not work.$

$Substituting $x=-7$ into (2) gives us $3\times 49-14m+15 = 0\Rightarrow m = \frac{81}{7}$.$

$Then substituting $x=-7$ into the L.H.S. of (1) with $m=\frac{81}{7}$ gives$

$\begin{align*}-7^3+7^2 \times \left( \frac{81}{7} \right) -15\times 7-7 &= 112\neq 0 \end{align*}$

$so Equation (1) is not satisfied, so a double root of $\alpha =-7$ does not work.$

$In conclusion, $m=-9$ with a double root of 1 is the one and only solution to the problem. So the answer is \boxed{$m=-9$}.$

Polynomials (roots) (1 Viewer)

senarillosk

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InteGrand

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