CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$\alpha $$ and $$ \beta$$ use the zeroes of the polynomial $$x^2 + px + q$$, then find the value of $$\left[\dfrac{\alpha}{\beta} + 2\right] \times \left[\dfrac{\beta}{\alpha} + 2 \right]$$.


Solution

Given equation: $$x^2+px+q$$

So, $$a=1, b=p$$ and $$c=q$$

$$\left [ \dfrac{\alpha }{\alpha }+2 \right ]\left [ \dfrac{\beta }{\alpha }+2 \right ]$$

$$=1+2\left ( \dfrac{\alpha ^2+\beta ^2}{\alpha \beta } \right )+4$$

$$=2\left ( \dfrac{\left (\alpha +\beta \right )^2-2\alpha \beta}{\alpha \beta } \right )+5$$

$$\alpha +\beta=\dfrac{-b}{a}, \alpha \beta=\dfrac{c}{a}$$

$$=2\left ( \dfrac{\left (\dfrac{-b}{a} \right )^2-2\dfrac{c}{a}}{\dfrac{c}{a}} \right )+5$$

$$=2\left ( \dfrac{\dfrac{b^2-2ac}{a^2}}{\dfrac{c}{a}} \right )+5$$

$$=2\left ( {\dfrac{b^2-2ac}{ac}} \right )+5$$

$$=2\left ( {\dfrac{b^2}{ac}}-2 \right )+5$$

Since, $$a=1, b=p$$ and $$c=q$$

$$\Rightarrow \left [ \dfrac{\alpha }{\alpha }+2 \right ]\left [ \dfrac{\beta }{\alpha }+2 \right ]=2\left ( {\dfrac{p^2}{q}}-2 \right )+5$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image