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Determine the value of $a$ such that $x^2-11x+a=0$ and $x^2-14x+2a=0$ may have a common root.

My attempt:

Let the common root be $\alpha$

On substituting $\alpha$ in both equations and then subtracting, $a = -3\alpha$

How do I continue from here? What are the other conditions for them to have common roots?

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  • $\begingroup$ The resultant (en.wikipedia.org/wiki/Resultant) would give a direct solution of the problem. In this case, the resultant of the two polynomials is $a^2 - 24a$ so they have a common root if and only if $a = 0$ or $a = 24$. $\endgroup$ – Daniel Schepler Sep 4 '17 at 16:04
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Eliminate $a$. Then, $$-2a=2(x^2-11x)=x^2-14x$$ $$x^2-8x=0$$ Therefore, the common root should be either $0$ or $8$.

If the common root is $0$, then $a=0$.

If the common root is $8$, then $a=24$.

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Another way to get $a$ directly, without first determining the common root.

  • Eliminating $x^2$ between the equations by direct subtraction gives $3x-a=0 \iff x = \frac{a}{3}\,$.

  • Eliminating $x$ between the equations by multiplying the first one by $14$, the second one by $11$ then subtracting the two gives $3x^2-8a=0 \iff x^2 = \frac{8a}{3}$.

Equating $(x)^2=x^2$ between the two previous expressions gives $\frac{a^2}{9}=\frac{8a}{3} \iff a \in \mathbb\{0, 24\}$.

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If you do the calculation correctly $a=3\alpha$. So,$x^2-11*x+3\alpha=0$ is the first equation and it has root $\alpha$. So $\alpha^2-11*\alpha+3\alpha=0$, that is $\alpha(\alpha-8)=0$. So, $\alpha=0,8$. If common root $\alpha=8$ then $a=3\alpha=3*8=24$. If $\alpha=0$ then $a=0$.

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  • $\begingroup$ But OP wants the value of $a$. $\endgroup$ – John Wayland Bales Sep 4 '17 at 4:56
  • $\begingroup$ I have edited the answer. $\endgroup$ – Arindam Sep 4 '17 at 5:08

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