Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If p and q are non-zero real numbers and <img alt="\alpha ^{3}+\beta ^{3}=\; -p \; ,\; \alpha \beta =q" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Calpha%20%5E%7B3%7D&plus;%5Cbe

If p and q are non-zero real numbers and \alpha ^{3}+\beta ^{3}=\; -p \; ,\; \alpha \beta =q ,  then a quadratic equation whose roots are   \frac{\alpha ^{2}}{\beta } ,\; \frac{\beta^{2} }{\alpha }    is :

Option: 1

px^{2}-qx+p^{2}=0          


Option: 2

qx^{2}+px+q^{2}=0


Option: 3

px^{2}+qx+p^{2}=0


Option: 4

qx^{2}-px+q^{2}=0


Answers (1)

best_answer

\\\text{Given } \alpha^{3}+\beta^{3}=-p\text{ and }\alpha \beta=q\\\text{ Let } \frac{\alpha^{2}}{\beta}\text { and } \frac{\beta^{2}}{\alpha}\text{ be the root of required quadratic equation.}

\\\text { So, sum of roots = } \frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}=\frac{-p}{q} \\ \text { and product of roots = } \frac{\alpha^{2}}{\beta} \times \frac{\beta^{2}}{\alpha}=\alpha \beta= q

Hence, the required quadratic equation is 

\\x^{2}-\left(\frac{-p}{q}\right) x+q=0 \\ \Rightarrow x^{2}+\frac{p}{q} x+q=0 \\\Rightarrow q x^{2}+p x+q^{2}=0

Posted by

jitender.kumar

View full answer

Similar Questions

Latest Question