Question

quadratic equation

Answer 1

did you miss an x out of the second term?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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))

Answer 6

i tried that and got

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

Answer 7

answer given is \( c.)\)

Answer 8

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

Answer 9

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 ?

Answer 10

That is clever!

Answer 11

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

Answer 12

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

Answer 13

@phi