IIFT 2018 Question 101

Question 101

The roots of quadratic equation $$y^2 -8y + 14$$ are $$\alpha$$ and $$\beta$$. Find the value of $$(1 + \alpha + \beta^2)(1 + \beta + \alpha^2)$$

Solution

$$\alpha\beta\ \ =\ 14$$

$$\alpha\ +\beta\ \ =\ 8$$

$$(1 + \alpha + \beta^2)(1 + \beta + \alpha^2)$$ = $$1+\beta\ +\alpha\ ^2+\alpha\ +\alpha\ \beta\ +\alpha\ ^3+\beta\ ^2+\beta\ ^3+\left(\alpha\ \beta\ \right)^2$$

 $$\alpha\ ^2+\beta\ ^2=\left(\alpha\ +\beta\ \right)^2-2\alpha\ \beta\ \ =\ 64-28\ =\ 36$$

$$\alpha\ ^3+\beta\ ^3=\left(\alpha+\beta\ \ \right)^3-3\alpha\ \beta\ \left(\alpha\ +\beta\ \right)=512-336=176$$

Hence $$1+\beta\ +\alpha\ ^2+\alpha\ +\alpha\ \beta\ +\alpha\ ^3+\beta\ ^2+\beta\ ^3+\left(\alpha\ \beta\ \right)^2$$ = 1 + 8 + 36+176+14 + 196 = 431


View Video Solution


Create a FREE account and get:

  • All Quant Formulas and shortcuts PDF
  • 40+ previous papers with solutions PDF
  • Top 500 MBA exam Solved Questions for Free

Comments

Register with

OR

Boost your Prep!

Download App