## mathmath333 one year ago quadratic equation

1. MrNood

did you miss an x out of the second term?

2. ganeshie8

are you sure there isn't a "x" missing in the middle term ?

3. mathmath333

\large \color{black}{\begin{align} &\normalsize \text{If the roots of the equation }\hspace{.33em}\\~\\ &a(b-c)x^2+b(c-a)x+c(a-b)=0 \hspace{.33em}\\~\\ &\normalsize \text{are equal , then }\ a,b,c \ \text{are in} \hspace{.33em}\\~\\ &a.)\ AP \hspace{.33em}\\~\\ &b.)\ GP \hspace{.33em}\\~\\ &c.)\ HP \hspace{.33em}\\~\\ &d.)\ \normalsize \text{cannot be determined} \hspace{.33em}\\~\\ \end{align}}

4. ganeshie8

you may simply set the discriminant equal to 0 and try to get a relation between a,b,c

5. MrNood

but the discriminant contains a,b & c and is only one equation not sure you can derive much from that b^2 = 4ac (but NOTE this refers to the standard quadratic form - NOT the abc in your equation))

6. mathmath333

i tried that and got \large \color{black}{\begin{align} b^2(a^2+c^2)-2abc(a+c-2b)=0 \hspace{.33em}\\~\\ \end{align}}

7. mathmath333

answer given is $$c.)$$

8. mathmath333

so i have to show \large \color{black}{\begin{align} &b^2(a^2+c^2)-2abc(a+c-2b)=0\hspace{.33em}\\~\\ &\Longleftrightarrow\hspace{.33em}\\~\\ &\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \hspace{.33em}\\~\\ \end{align}}

9. amoodarya

hint : if a+b+c=0 in standard $ax^2+bx+c=0 \\x_1=1 ,x_2=\frac{c}{a}$ in your case sum of coefficient is =0 a(b−c)+b(c−a)+c(a−b)=ab-ac +bc -ab +ac -bc =0 can you go on ?

10. ganeshie8

That is clever!

11. mathmath333

i am confused , how u got $$x_1=1$$ and $$x_2=\dfrac{c}{a}$$ if $$a+b+c=0$$

12. MrNood

I would also dispute that a+b+c=0 implies that ax^2 + bx +c =0

13. mathmath333

@phi