Math, asked by nunknown984, 3 months ago

find zeroes of p(x) = abx²+ (b²-ac)x - bc. Using alpha and beta​

Answers

Answered by rkcomp31
2

Answer:

\bf The \ roots \ are :\\\\\frac ba \ and  \  \frac{-c}{b}

Step-by-step explanation:

\bf p(x)=abx^2+(b^2-ac)x-bc\\\\=x^2 +( \frac{b^2}{ab} -\frac{ac}{ab} )x-\frac{bc}{ab} \\\\=x^2 +( \frac{b}{a} +\frac{-c}{b} )x+\frac{b}{a}\times\frac{-c}{b} \\\\Now let \ \alpha = \frac ba \ and \  \beta =\frac{-c}{b} \\\\Then \ p(x)=x^2-(\alpha +\beta)x+ \alpha \beta \\\\Thus \ the \ roots \ are :\\\\\frac ba \ and  \  \frac{-c}{b}

Answered by vaishubh1707
5

Answer:

Refer attached image for your answer.

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