Question

If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( p(x)=4 x^{2}+3 x+7 \), then find the value of: 

(a) \( \frac{1}{\alpha}+\frac{1}{\beta} \)

(b) \( \alpha^{2}+\beta^{2}-3 \alpha \beta \) 

(c) \( \frac{4}{\alpha}+\frac{4}{\beta} \) 

(d) \( \left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+6 \alpha \beta \)

Answer 1

α + ß = -3/4

αß = 7/4

(a) \(\frac{α+ß}{αß}=\frac{-3}7\)

⇒ \(\frac1\alpha+\frac1{\beta}\) = \(\frac{-3}7\) 

(b) α² + ß² - 3αß = (α + ß)² - 5αß

 = \((\frac{-3}4)^2 - 5\times\frac74\) 

 = \(\frac9{16}-\frac{35}4\) 

 = \(\frac{9-140}4\) = \(\frac{-131}4\)

(c) \(\frac4{\alpha}+\frac4{\beta}=\frac{-12}4\) (From (a))

(d) \((\frac{\alpha}{\beta}+\frac{\beta}{\alpha})+6\alpha\beta\) 

\(=\frac{\alpha^2+\beta^2 +2\alpha\beta-2\alpha \beta}{\alpha\beta}+6\alpha\beta\)

\(=\frac{(\alpha^2+\beta^2)^2-2\alpha\beta}{\alpha\beta}+6\alpha\beta\) 

\(=\cfrac{(-\frac34)^2-2\times\frac74}{\frac74}+6\times\frac74\)

\(=\cfrac{\frac{9-56}{16}}{\frac74}+\frac{21}2\) 

\(=\frac{-47}{7\times4}+\frac{21}2\)

\(=\frac{-47+294}{28}\)\(=\frac{247}{28}\)