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mathmath333

  • one year ago

quadratic equation

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  1. MrNood
    • one year ago
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    did you miss an x out of the second term?

  2. ganeshie8
    • one year ago
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    are you sure there isn't a "x" missing in the middle term ?

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\normalsize \text{If the roots of the equation }\hspace{.33em}\\~\\ &a(b-c)x^2+b(c-a)x+c(a-b)=0 \hspace{.33em}\\~\\ &\normalsize \text{are equal , then }\ a,b,c \ \text{are in} \hspace{.33em}\\~\\ &a.)\ AP \hspace{.33em}\\~\\ &b.)\ GP \hspace{.33em}\\~\\ &c.)\ HP \hspace{.33em}\\~\\ &d.)\ \normalsize \text{cannot be determined} \hspace{.33em}\\~\\ \end{align}}\)

  4. ganeshie8
    • one year ago
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    you may simply set the discriminant equal to 0 and try to get a relation between a,b,c

  5. MrNood
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    answer given is \( c.)\)

  8. mathmath333
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    That is clever!

  11. mathmath333
    • one year ago
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    i am confused , how u got \(x_1=1\) and \(x_2=\dfrac{c}{a}\) if \(a+b+c=0\)

  12. MrNood
    • one year ago
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    I would also dispute that a+b+c=0 implies that ax^2 + bx +c =0

  13. mathmath333
    • one year ago
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    @phi

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