Common root question on quadratics equations to show that $a+b+c=0$

Question

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

I tried this by thinking that $\alpha$ is the common root and then I got by substituting and solving ,

$(b^2+c^2)(b-c)+a(b+c)^2=0$

how can I proceed from here ? Any ideas ?

Answer 1

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4\neq 0$.

Answer 2

Hint: the roots of $f$ are $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

Answer 3

Hint 2: An alternative approach is to say that

$$ f(x) = a (x - u) (x - v) $$ for some numbers $u$ and $v$ (the roots of $f$), and $$ g(x) = a(x-u) (x - w) $$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

Answer 4

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-\frac ba,\quad pq=-\frac ca$$ and $$p+r=-\frac ca,\quad pr=\frac ba$$

Answer 5

It's wrong.

Try $b=c=0$ and $a=1$.

Answer 6

$$f(x)=ax^2+bx-c$$ $$g(x)=ax^2+cx+b$$ for f(x), the roots are given by: $$(1)\,x=\frac{-b+\sqrt{b^2+4ac}}{2a} \text{and}\, x=\frac{-b-\sqrt{b^2+4ac}}{2a}$$ for g(x), the roots are given by: $$(2)\,x=\frac{-c+\sqrt{c^2-4ab}}{2a} \text{and}\,x=\frac{-c-\sqrt{c^2-4ab}}{2a}$$ since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

Answer 7

Counterexample: $a=c, b=0$. That is: $$f(x)=ax^2+bx-c=ax^2-a=0 \Rightarrow x=\pm 1;\\ g(x)=ax^2+cx+b=ax^2+ax=0 \Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then: $$f(x)= ax^2+bx-c=0 \iff ax^2+bx+a+b=0 \ \ \ \ \ \ (1) \\ g(x)=ax^2+cx+b=0 \iff ax^2-ax-bx+b=0 \iff (x-1)(ax+a-b)=0 \ \ \ \ (2) $$ Then, from $(2)$: $$x-1=0 \stackrel{(1)}{\Rightarrow} a=-b.\\ ax+a-b=0 \Rightarrow x=\frac{b-a}{a} \Rightarrow a\frac{(b-a)^2}{a^2}+b\frac{b-a}{a}+a+b=0 \Rightarrow a^2+b^2=ab.$$

Answer 8

Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since $$ f(x) - g(x) = (b-c)x -(b+c), $$ then common root should be $$ r = \frac {b+c} {b-c} \quad [b \neq c]. $$ Plug this into $f(x)$: $$ a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0, $$ which is $$ a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0. $$ Hence $$ a + b + c = \frac 1 {(b+c)^2} (2c^3 + 4b^2c + 2bc^2) = \frac {2c} {(b+c)^2} (c^2 + 2b^2 + bc). $$ If $c = 0$, and if $b\neq 0$, then $a+b + c = 0$. If $c \neq 0$ and $b +c \neq 0$, then $$ |a +b +c| = \frac {2|c|} {(b+c)^2} \left( \left( c+\frac b2\right)^2 +\frac 74 b^2\right) > 0. $$ Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then $$ f(x) = x^2 - 1= (x+1)(x-1),\quad g(x)= x^2+x =x(x+1), $$ clearly they have a common root $-1$ but $a+b+c = g(1)=2\neq 0$.

Conclusion: such claim fails.

Answer 9

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b \ne c$, then $f(x)=ax^2+bx \color{red}{+}c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $\,(b-c)(x-1)=0\,$, therefore the common root must be $\,x=1\,$.