2

The number of roots common between the two equations

$x^3+3x^2+4x+7=0$ and $x^3+2x^2+7x+5=0$ is

$\color{green}{a.)\ 0 } \\~\\ b.)\ 1 \\~\\ c.)\ 2 \\~\\ d.)\ 3 \\~\\ $

i tried to solve both equations by subtracting then

$x^3+3x^2+4x+7-(x^3+2x^2+7x+5)=0 \\ x^2-3x+2=0 \\ x=2, \ 1$

but the answer is given as option $a.)$

I look for a short and simple way.

I have studied maths up to $12$th grade. Thanks!

Martin Sleziak
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R K
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3 Answers3

5

You've found the $x$ values where the two expressions are equal. However, at neither of these $x$-values are the expressions equal to $0$, which is what you need for roots.

paw88789
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2

You can apply the Euclidean algorithm to the polynomials to find they're coprime, hence they can't have a common root (if they had a common root, say $\alpha$, $x-\alpha$ would be a common factor).

Bernard
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  • can u give me a link on how to use that Euclidean algorithm algorithm – R K Jun 20 '15 at 15:14
  • I gave a detailed explanation with an example in the case of integers in this answer. The recurrence relations are the same for polynomials, so I hope it will be clear enough. – Bernard Jun 20 '15 at 16:10
1

Considering $$f(x)=x^3+3x^2+4x+7$$ $$g(x)=x^3+2x^2+7x+5$$ you could notice that their derivatives never cancel in the real domain. So, $f(x)=0$ has only one real root and same for $g(x)=0$. So, the maximum number of common roots is $1$.

Now, inspection :

  • $f(-3)=-5$, $f(-2)=3$; so the root for $f(x)=0$ is somewhere between $-3$ and $-2$.
  • $g(-1)=-1$, $g(0)=5$; so the root for $g(x)=0$ is somewhere between $-1$ and $0$.

So, no common root.

Claude Leibovici
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  • i didnt understand what u mean by this line "you could notice that their derivatives never cancel in the real domain" . i never really learned higher calculus – R K Jun 20 '15 at 15:13
  • Since the derivative does not cancel in the real domain, the function does not go through any maximum or minimum. Since the functions are cubic polynomial with postive coefficient for $x^3$, the curves go from $-\infty$ to $+\infty$; so, they only have one real root. Is this better ? Just post. Cheers :-) – Claude Leibovici Jun 20 '15 at 15:52
  • can u please show me algebraically why derivatives never cancel in real domain. – R K Jun 21 '15 at 09:32
  • $f'(x)=3x^2+6x+4$, $g'(x)=3x^2+4x+7$. These are quadratic; compute the famous $b^2-4ac$ : it is negative for both of them. So no real root for $f'(x)=0$ and for $g'(x)=0$. Then no maximum and no minimum; the only one real root for $f(x)=0$ and for $g(x)=0$. Is this better ? – Claude Leibovici Jun 21 '15 at 09:36